precalculus

[whole perch]

they think im those comp math nerds

whats ur GPA

[tranquil inlet]

4.0 straight a's but physically cannot take ap classes (in freshamn)

[whole perch]

oh

no honors?

or is that unweightes

[tranquil inlet]

our school doesnt weight honors

😭

[whole perch]

oof

[tranquil inlet]

what aps u taking next year

[whole perch]

ap calc ab/bc (im gonna request to change my course load to ap calc ab and ap pysc instead tho), ap us gov, ap csp

[tranquil inlet]

i see

im taking ap chem, ap calc, and ap cs

[whole perch]

how ap chem?

[tranquil inlet]

its what all the sophomores take

[whole perch]

oh they only let us take normal chem

or honors

im taking honors

[tranquil inlet]

nah my school u can take regular honors or ap chem but if u fck up, u fck up

taking summer school honors chem tho

[whole perch]

oh

alr bro im sleepy af

cya

gl

[tranquil inlet]

ok bye

[whole perch]

with life

[tranquil inlet]

you 2

[whole perch]

bye

[tranquil inlet]

byee

[hallow lodge]

how is -2.087 greater than 0.05

is it going by the graph

[pure sable]

hiii can someone help me?

pleaseee

how would i graph this if the radius has a square root?

[fiery creek]

what are you trying to do here

[pure sable]

graph circle

i think i figured it out i just have to approximate

[fiery creek]

are you graphing a circle at center (4, 1) with radius 2 sqrt(2)?

[pure sable]

yes

[fiery creek]

2 sqrt(2) is just a constant

you would graph it the same way as anything else

[pure sable]

ohhhh ok

thank you 🙂

[fiery creek]

except it's kinda pointless to do it by hand since there won't be much accuracy

[fiery creek]

np lil dawg

[pure sable]

yea im forced to

or else she’ll give me a 0

[fiery creek]

your teacher?

[willow bear]

i mean there are grid points at distance 2sqrt(2) from (4,1)

so at least you can get those exactly

and eyeball the rest

2 sqrt(2) is a little bit less than 3

[pure sable]

yes

no graph no credit

[compact spade]

How would I go about doing these

I can solve 11-18 but not 5-10

[proven void]

5. C(3,1)

no?

[proven void]

does 5 c(3,1) here?

also this is not a question for #probability-statistics ?

why precalculus?

[compact spade]

Idk, I'm taking precalc honors and this is a hw assignment from that class

Could I uh

Get some help with these

Still a little lost :(

[jovial coral]

Sure let me read them

[compact spade]

Thank you 😭

I need to go to bed I have a sleep schedule to fix

[proven void]

6 is a combi?

[jovial coral]

Lemme see

Yes I think so

[proven void]

doesnt make any sense that the order matters

c(1,8)

[proven void]

for permutations order matters i believe

[jovial coral]

Yes

[proven void]

7 is combi, c(1,3) aswell

order doesnt matter

[jovial coral]

I need to go to sleep. Sorry I couldn't help more. gn

[proven void]

gn sleep well scraxzt

[compact spade]

Its ok, sleep well

[jovial coral]

Thanks

[compact spade]

@willow bear I'm sorry for pinging you - if you'd prefer that I don't just lmk and I won't do it again but I'm so lost on this hw, I need to go to bed, and I really, really need help

[willow bear]

which one(s)

[compact spade]

5-10

[willow bear]

these?

ok, do you have progress on any one of these? if you do, share your progress and we'll start with that. otherwise we will go through them in order.

[compact spade]

I mean, I know the functions for permutations and combinations

I know that permutations means order matters

I just don't know how to determine which is which

[willow bear]

im talking about these specific problems -- do you or do you not have progress on at least one of them?

[compact spade]

No, I don't

[willow bear]

ok

for some of these you dont need to know permutations/combinations and just need to know shit like the multiplication principle

which may go by many names or none at all

but let me give you a different setup as an example

have you ever seen, or held in your hands, a deck of cards? @compact spade

[compact spade]

Yes

[willow bear]

like regular poker cards

ok

so you know that the cards come in 4 suits and within each suit there are 13 ranks

[compact spade]

Yes

[willow bear]

and there's exactly one card for each suit-rank combination

for a total of 4 * 13 = 52 cards

the same principle applies to problem 5, in which you choose from 3 different albums and 2 different formats (the choice counts being independent of each other)

do you see what i'm talking about here?

[compact spade]

Yes!

[willow bear]

ok great

problems 5, 6, 8 and 10 can be done with this method more or less directly

we'll come back to 7 and 9 in a moment

[compact spade]

Ok, perfect

Thank you so much

I'll let you know when I'm onto 7 & 9

[willow bear]

ok sure

you can ping me when that happens

(but you will have to remind me cause i might forget this)

(like just tell me the problems you're following up on or w/e)

[compact spade]

Onto 9! We can do 7 also but I figured that one out by just writing them down

@willow bear

[willow bear]

alright

so for 7 you can also apply the multiplication principle in a similar way

carlos has:

• 3 subjects to choose from as his first
• 2 subjects to choose from as his second (because one has already been done as the first)
• 1 subject to choose from as his last

for 3 * 2 * 1 = 3! = 6 ways

for 9 proceed in a similar fashion but place the dictionary first

[compact spade]

ohhhhhh ok

that's really easy thank you

can i like

pay you??

you've helped me with my math homework like 5 times now and i feel like i should do something back idk

[willow bear]

i mean uh

all the shit i do on here i do pro bono

if any money were to change hands we would not want to do it on here bc rules

[compact spade]

that's true

[dense coyote]

hey sorry for replying to this message but is PNC tough ? we have this on our syllabus & our sir told it will be done at end since rarely qs come from it

[willow bear]

sorry for replying to this message
could have just pinged me

anyway all toughness is relative

so i cant answer that

[viscid thistle]

Depends on how deep your syllabus is. Once you understand where to use either and practise questions, it becomes fairly trivial. In cases like special arrangements on chessboards, dearrangements etc., it becomes quite demanding.

[dense coyote]

thnx, our syllabus is pretty simple but the qs are made quite tough to check the aptitude

[agile sparrow]

Jee right

[dense coyote]

no CAT, JEE is way beyond this 🙂

[quiet bane]

Hi

Is anyone here good at sequences

[winter comet]

people in calc chat prob even better ngl

you should prob just post ur problem tho

[quiet bane]

I tried several time no one helps i have to do it before tomorrow and i have to sleep its literally 2:00 AM

[winter comet]

where did you post the problem?

[quiet bane]

#help-44｜stanley-🌲-v2-dans

Here

And in disscution / prealg / geometry

The problem is that the exercise isn't that hard

[winter comet]

try to stick to just one channel

and uh yeah i dont know series that well

[fallow field]

Is anyone free to help me study chapter 3

3.1 to 3.5. I have a study guide.

[hushed sphinx]

You still haven't told anyone which book.

[fallow field]

102 Lecture Notes

[coarse ginkgo]

Can someone please help me visualize this. I’m not sure if i did it correctly

[white rapids]

Can someone explain why x is 16 even though k = 1/2

Shouldnt the x value be 4 since the original is 8?

I did the graph on desmos, but graph g seems to be compressed

Someone explain pls

Nvm

[shut monolith]

this would be a correct visualization (im taking the lighthouse to be a single point as we don’t know about any internal details regarding the lighthouse at all)

[viscid thistle]

Hey y'all, I don't take pre-calc, but I am currently teaching myself as it was something that I "skipped" over. I have had mixed opinions on skipping pre-calc and going straight to calc 1. Looking for some opinions here, anything is appreciated!

[whole perch]

yo i was using my calculator on the test

and it said -1^2

was -1

am i buggin

or what

how

shouldn'[t it be 1

[river drift]

by convention, we define the order of operations to say that
[ -1^2 = -(1^2) \ne (-1)^2 ]

[obsidian monolith]

cloud

[whole perch]

oh

shit

iok

[worn bridge]

HOW DO I TRANSLATE THIS to summation notation
If i had a penny that doubled each time per day, how many would i get in 30 days

[muted obsidian]

2^30 ?

[echo sinew]

1 + 2 + 2² + 2³ + ... + 2³⁰

This is the closed formula

In your case it's 1•(1 - 2^31)/(1-2), which is 2³¹ - 1

[willow bear]

op admitted that the problem is one he made up himself

but also as-written it doesn't suggest that sum

we have a magical penny that clones itself every night, not a money machine that starts out paying you a penny a day and then pays twice as much every day as it did the last

[echo sinew]

Oh you're right, I must have misread it

Then prolly the answer they want is 2³⁰

[daring tapir]

am i just too dumb or should the answer he wants be 2^29 instead?

[willow bear]

the problem is stated too vaguely

[daring tapir]

true

[worn bridge]

Perhaps I should revise the question, so, If I had a penny (0.01$) that doubled each day for 30 days, by day 2 (0.02), by day 3(0.04), next(0.08), and so on, so, how much would I get in 30 days? In this time in terms of VALUE, not the quantity of pennies. How would i put it in summation notation form?

[willow bear]

why do you specifically need this "in summation notation form"?

you yourself said the question is made up by you.

"summation notation" is not a panacea, and is not a thing you should be trying to shoehorn into every single problem you see or think of

anyway, with this more precise statement, your penny undergoes 29 doublings, so you end up with 2^29 = 536870912 pennies, or $5,368,709.12 at the end of it.

[hallow lodge]

i need help with factoring

like literally simple factoring

how do u do it

[willow bear]

https://youtu.be/9MUDAr1PnlE?feature=shared

first hit on youtube search for "ochem tutor simple factoring"

[hallow lodge]

yt videos js show the factored function automatically

[willow bear]

do you have a particular problem or problems that you are having a hard time with?

[hallow lodge]

could u tell me the formula for it

[willow bear]

1. avoiding the question
2. there is no "formula"

[hallow lodge]

having problem with it in general

what abt the steps

ive been having problems with factoring since 1st sem

[uncut mulch]

post specific problems you have an issue with

[winter comet]

4x^2 + 8x
= 2 * 2* x * x + 2 * 2 * 2 * x
you're multiplying all these factors, if you are adding two terms then you can take common factors out of the two terms
you can see that both terms have two 2s and one x in common, so you can "factor" these out of both terms.
you can take out, 2 * 2 * x, and the remainder is (x + 2)
so, it equals
2 * 2 * x * (x+2)
= 4x(x+2)
this ^ is the factored form, because it took all the common factors in each term out, and now you're multiplying by whatevers left

[tranquil inlet]

How would I set this up to prove with induction

[shut monolith]

are you proving for all numbers {0,1,2,3,…} or {1,2,3,4,…}?

[hushed sphinx]

Divide into cases according to whether n is 3k or 3k+1 or 3k-1 for some k.

Oops, didn't see you wanted induction.

[worn bridge]

There is no formula, but, thetre is a first step to factkring which is takibg the gcf of all terms, so say, 4x+2x has (2x) as the gcf, so if you ffactor it, you will get 2x(2+1) basically you reverse foil method

[tranquil inlet]

Does it really matter 0+0=0

[shut monolith]

The question asks to “prove n^3 + 2n is divisible by 3”, but it’s not clear what “n” is

[hushed sphinx]

It's true for all n in Z.

[pure sable]

hi can someone help me because i wasn’t here for math today

[pure sable]

what is foci?

[shut monolith]

so the center, vertices, covertices, and foci (plural for focus) are all points, so your answer for each of them will be ordered pairs

i gtg sorry someone else should be able to explain hopefully

[pure sable]

ur good thanks

[golden cairn]

How do I solve this? I have been trying for like 30 minutes but I don’t even know

[winter comet]

try taking log of both sides

[golden cairn]

I already did

What do I do after I get: (3x)log6=(2x-3)log2

I tried dividing by log 6 on both sides but I don’t know

[winter comet]

yes, you can do that

remember log6 and log2 are both just constants

[golden cairn]

Ok

Then what do I do after that?

[winter comet]

factor it out

it might have been better if you divided by log2 instead but what ur doing works as well

[golden cairn]

Factor what out?

[winter comet]

sorry i meant distribute 💀

i don't know why i say factor instead of distribute 💀

i can't think of the word distribute lol

[shut monolith]

is there any part of this you still need assistance with?

[pure sable]

no but i still need help with this problem

like how to graph vertices if one is a square root

[shut monolith]

since the instructions just say “sketch”, you should be fine to estimate, just treat sqrt(12) like it’s 3.5, since that’s what it is approximately

[pure sable]

thanks so much

[stiff kelp]

factoring is ez

u need help

[winter comet]

why you think hes here genius 🗿

[finite root]

Can you help me with this?

[winter comet]

do you know the cos(a+b) identity

[finite root]

yes

[winter comet]

can you write it out?

[finite root]

cosacosB-sinasinB

[willow bear]

ok, so make a table:

\begin{tabular}{r||c|c}
&$\alpha$&$\beta$\
\hline\hline
cos & & \
\hline
sin & &
\end{tabular}

[obsidian monolith]

|Ann⟩

[willow bear]

fill it out based on the given data

bc you're gonna need all four numbers in it to find cos(α) cos(β) - sin(α) sin(β)

[devout bison]

oh ik it

[willow bear]

@devout bison so you want someone to give you a worksheet of basic summation notation problems?

[devout bison]

?

[willow bear]

(cont'd from #multivariable-calculus )

[devout bison]

this is cos(a+b)

[willow bear]

you said you wanted some summation questions to practice, right?

[devout bison]

a is alpha b is betha

[devout bison]

yeah

[willow bear]

beta, not betha.

anyway ok gimme a couple more minutes.

[devout bison]

okay thx by the way

ig ik precalculus by the way

[silent oasis]

while finding the limit by direct substitution how does b/0 (where b≠0) denote an asymptote?

[willow skiff]

You can't divide by 0

[willow bear]

@devout bison here you go

[willow bear]

it means the function behaves similar to 1/x near your limit point

[silent oasis]

but 0/0 doesn't denote asymptote

[devout bison]

but u can in infinity

when k/x xgoes to infinity y goes to 0

[willow skiff]

0/0 is a special case
Sometimes it's a removeable discontinuity so the limit has to exist, and other times it's not going to have a limit at all cause it's not continuous

[willow skiff]

Yes

[willow skiff]

0/0 can still be an asymptote

it's just that it doesn't have to be one, it depends on the functions

[silent oasis]

what's 1/x here?

[willow skiff]

the function y = 1/x

the green one is like y = 1/x around x = 0

[willow bear]

@devout bison if you want you can DM me your work for any of these problems in the worksheet, but with two rules:

1. include the questions themselves in your DM.
2. show work and not only answers. i won't accept answers without work, even if they are right.

[devout bison]

is there a quick way for1)b)

[willow bear]

define quick?

[devout bison]

yeah kinda

instead of going long

[willow bear]

quick compared to what?

[devout bison]

1 +4+9 etc

[willow bear]

the sum in 1b) has length 4. i want you to do it longhand.

i have even instructed you as such.

also look more carefully at the notation.

[silent oasis]

got it. thanks.

[willow skiff]

np

[willow bear]

@devout bison i made the sums in Q1 short on purpose, so that working them out in some kind of "smart" way would take more effort than just doing as i instructed.

[devout bison]

okay

q2 is 1 ig

q1)a)70

the number on the up means that u will go as much as that or until that

[willow bear]

work, not only answers.

[devout bison]

or until that

[devout bison]

indeed i used quick method

[willow bear]

you have to write out your work properly and show it

until then i will not accept any answers.

[devout bison]

i dont have a phone am on pc not sure if cam will see it

[willow bear]

my problems, my rules.

[willow bear]

as in you don't have a phone at all, or that you don't have it on your person at the moment?

[devout bison]

at all

i broke it last time

[willow bear]

bruh ok

well, you're gonna have to figure out how to write it nicely.

cause im still not gonna accept answers without work.

[devout bison]

can u help me out on 4th @willow skiff

@willow bear

a quick method

wait

exactly like the graphics right

[willow bear]

idk what you mean by "the graphics"

[devout bison]

functions

but we only take complete numbers here

and thats a big problem

[willow bear]

that explains precisely nothing...

idk what on earth you are trying to say there sorry

are you trying to ask me something, or are you trying to bounce an idea off of me?

[devout bison]

can u pls give me a certain method that i could apply in all of these

[willow bear]

all 4 questions?

[devout bison]

no just a method for all summations

especially 4th

1 and 3 were the same and method written

[willow bear]

you want a panacea

[devout bison]

?

[willow bear]

what you're asking me for is a "solve everything" method

such a method does not exist

[devout bison]

solve every summation

?

for example in functions there is

1 method we apply to find areal of all these

okay so how will 4th be

@willow skiff

<@&286206848099549185>

@willow bear

[devout bison]

how will 4th be

[willow bear]

work out the first few terms, and see if you notice a pattern.

[devout bison]

nope there wasnt

i did all

except for

1)c)

and 4th

@everyone do u know a certain method to find out summation

a method that could be applied for all summations

[willow bear]

don't ping everyone. it's incredibly rude.

[willow bear]

ok then write out the first 8 terms of the summation $\sum_{k=0}^{85} \cos(k\pi/2)$ right here and now. not their sum, but the first 8 terms themselves.

[obsidian monolith]

|Ann⟩

[devout bison]

0 1 2 3 45 6 7 8

cos0

cos90

oh i see

it is cos90k

which is either 0 0r 1 0r -1

and they keep on each other

[willow bear]

no. 0, 1, 2, 3, 45, 6, 7 and 8 are not the first eight terms of my summation.

[devout bison]

4)17 ig

thanks to the Almighty

and what about 1)c

[willow bear]

wrong answer, and it is unsupported by work.

[devout bison]

why wrong

[willow bear]

(for future answers that you don't give work for, i will simply say "no work" no matter if you get it right or not.)

[willow bear]

impossible to tell without working.

[devout bison]

i worked in my mind

[willow bear]

not acceptable.

[devout bison]

why to write down on paper

by the way u won't even see what i write

and

yeah i see

it was wrong

[willow bear]

make a word document and write your work there. it has an equation editor.

and send screenshots of that.

[devout bison]

the answer is

1

on 4th

it is 1

[willow bear]

answers without working are not accepted.

[devout bison]

they roll into same after4th number

and when we reach 85th

they rolled

but one still left and that is 85pi*k/2=1

oh wait

[willow bear]

you are expressing yourself poorly.

[devout bison]

that is 0 either

[willow bear]

what is your native language?

[devout bison]

could u be more respective

what u have as a knowledge doesnt give u the right of boasting on others because

nothing we have is actually gained by our own

it is a gift

[willow bear]

i am not boasting on anybody.

[devout bison]

but u kinda acted rude on me

3 times

just how old r u

i am 16

[willow bear]

"you are expressing yourself poorly" is the most polite way i know to convey that point.

i am 24, if that matters at all.

[devout bison]

yeah it matters becuz at that age don't u think it is more impolite to insult someone smaller than u

[willow bear]

i am not insulting you.

or you could tell me what i said that you took as an insult.

[devout bison]

ok sorry then for misunderstanding

yeah

[willow bear]

ok, then what was it?

[devout bison]

1

[willow bear]

and what are the other 2 messages which you took as rude?

you said there were 3

[devout bison]

i said u already that i dont have a phone to img

[devout bison]

and this is kinda so rude

[willow bear]

ok, so maybe i should have suggested the alternative of "make a word doc and type shit there with equation editor" then and there.

[devout bison]

yeah

[willow bear]

and i honestly don't see how "my problems, my rules" is rude. these are my problems that i made -- and i get to dictate what i want to see in response to them.

if you don't like that, you don't have to do my problems.

[devout bison]

then can u say like

[willow bear]

i am not forcing you.

[devout bison]

i wanna see how uu do it

[willow bear]

you want me to see how i do #4?

[devout bison]

yeah

no

i did 4th

i only need a way for 3)c and and 1)c

by the way i already did 4th and explained the way i did upwards

i wanna see how uu do it- i typed it that u could talk like that since it is kinda more soft u see

[willow bear]

for 1c, just as for 1a and 1b, write the thing out longhand. it is only 5 terms.

[devout bison]

okay

[willow bear]

i am still interested in knowing what your native language is

there is a chance we can communicate better in your language

[devout bison]

well it is turkish

[willow bear]

ok, nevermind

[devout bison]

okay

yours is probably french ig?

or german

[willow bear]

no, my native language is Russian.

anyway here is 1c

[devout bison]

yeah but what about a quick method

for all equations in the universe

[willow bear]

such a method does not exist

[devout bison]

not equations but summations i wanted to say

[willow bear]

and still

such a method does not exist

[devout bison]

ups

okay so

[willow bear]

for 3c, my first step would be to split the summation like this: $$\sum_{n=0}^{15} n^2 + \sum_{n=0}^{15} 6n,$$ and also realize you can replace the starting index 0 with 1 in both summations (because $6\cdot 0=0$ and $0^2=0$)

[obsidian monolith]

|Ann⟩

[devout bison]

ohwait sorry

i mistyped

i don't need 3c

okay i did all

now i wanna ask u some questions

did u know something so called e^ix=cos (x)+i*sin(x)

what is the provement of this

?

@willow skiff @willow bear <@&286206848099549185>

i mean where did it come from

[willow bear]

the word is "proof", not "provement"

[willow bear]

stop pinging so many people so often.

it is 200 times more rude than anything i have done.

[devout bison]

okay

[willow bear]

anyway

[devout bison]

so

[willow bear]

one idea you could think about is that if you look at the function f(x) = cos(x) + i sin(x) [where we don't know what e^ix should mean yet], it satisfies the law f(x+y) = f(x) * f(y)

so it should behave like an exponential function, i.e. e^(kx) for some constant k

(because all exponential functions satisfy the same law)

if you take the derivative of our function, you will find f'(x) = -sin(x) + i cos(x) = i*f(x)

while the derivative of e^(kx) is k e^(kx), so k is the ratio of the function's derivative to itself

in our case this ratio will be i, so our function could be written as e^(ix)

(this is not rigorous. this is meant as one possible kind of intuition for euler's formula.)

(to make a fully rigorous proof, we would need to dig into the definitions of e^z, sin(z) and cos(z).)

[devout bison]

? my question

[willow bear]

what?

you asked about e^ix = cos(x) + i sin(x)

[devout bison]

yes

[willow bear]

i gave you my best attempt at something that explains why it is true, but without being heavy

bc i think you don't want the heavy shit

(it'll involve series, which are like summations but infinitely long -- and you are only beginning with finite ones!)

[devout bison]

sorry the message wasnt loaded so i typed again

f(x+y) = f(x) * f(y) is this always true?

[willow bear]

it is true for the function f(x) = cos(x) + i sin(x), and for all functions of the form f(x) = e^(kx).

[devout bison]

and here is something how can we draw these functions

with numbers that dont exist in reality

[willow bear]

"don't exist in reality" is almost a derogatory way to refer to complex numbers.

just because they aren't real does not mean they do not exist as mathematical objects.

[devout bison]

ik they exist in math

and a very

essential part

inclluding a miracle

the fractals like one u see in my prophile

[willow bear]

unfortunately, it is not easy to draw graphs of functions where the input or output (or both) is complex.

[devout bison]

so what we do

[willow bear]

for a function with real input and complex output, or complex input and real output, you can make a 3 dimensional sort of graph

with 2 axes for the complex variable and 1 for the real

it'll be a surface or a curve

but for complex->complex you would need 4 dimensions to do it like that

[devout bison]

so kinda like fractals but more complicated

[willow bear]

so usually for those functions, if they are plotted at all, they do something more sophisticated like colors

[willow bear]

nothing to do with fractals.

[devout bison]

u know how we draw fractals right?

[devout bison]

?

what do u mean by 4dimensions

and why did we use 2 axes for complex

[willow bear]

complex numbers form a plane

you know this, right?

[devout bison]

no

[willow bear]

bruh

you have a Mandelbrot set on your avatar and you talk about fractals but you don't even know basics of complex numbers?

[devout bison]

i only know what they are

i am mid school

and i searched how fractals work but idk what u say the complex surface

[willow bear]

ok well heres the issue

if you ask about stuff like euler's formula like you did above

people are going to assume you are familiar with some basics of complex numbers

so you should go learn that

[devout bison]

ik the euler number

and what it is

((x+1)/x)^x=y when x goes to infinity y goes to e

[devout bison]

ik basics

but not the "complex surface"

can u send a video orexplain it

[static mesa]

can i ask combinatorics here

[devout bison]

what is ur question

[static mesa]

jamies dad allows him to choose 8 cars for his birthday

jamie's dad has 6 aventadors , 5 r8s , 4 bmws

in how many ways can jamie pick such that there're an equal ammount of r8s and aventadors

@devout bison

[devout bison]

the amount of all possibilities

15P8

[static mesa]

order dont matter

i want the combinations only pls

[devout bison]

so i think the pssibility for each will be

6P1*5P1

6P2*5P2

6P3*5P3

6P4*5P4

6!*5!

<@&286206848099549185>

sum these

divide by

15P8

wanna get explanation

?

sum these- this will give u the n of possibilities where number of r8 and avendator equal

when u divide itwill give u the possibility

[static mesa]

uh can u explain the first part

[devout bison]

u see

u have 6 cars from that type

your ability of taking random u have 6 chances

u took one

u r left with 5 chances

one more take one minus chance

now for each single choice u can make on first run which is equal to 6

in the next run u have 5 different choices

but these willl change

it is like a timeline game

where u control the timeline

as u made a choice

u entered an alternate timeline where

u can make 5 different choices

with each next choice

u type that timeline more

but still

u have other timelines either

@everyone how can i put images there

now if u start with 5 cars

there are 5 possibilities

u made one

now u have 4 possibilities

but imagine

u could have made another choices

and ur next generation possibilities would have changed

if the order doesnt matter

instead of permutation we use combination

4P3 means

u start from 4

and each single time multiply it to -1 number

until the amount of numbers is 3

so 4P3=4 * 3 * 2

see 3 numbers

started from 4

and multiplyed them to each other

@static mesa

?

could u do it

also i made a mistake if u saw

i had to use the combination

so it will be

@static mesa

?

[static mesa]

oh yea

i think i get it

thx man

[devout bison]

so instead of p u will right C ig

there is only 1 difference between c and p

u divide p into the number of timeline generations(the number on the right)

this is a chat gpt explaining

i tought it will be more accurate so i sent this one

the previous was mine

[tender quest]

Please do not trust ChatGPT or similar AI tools for mathematical tasks, as they often generate output which "sounds correct" but has numerous factual or logical errors. Use of these AI tools to answer other people's help questions is strictly against server rules (see #rules).

[static mesa]

yup

[devout bison]

u divide p into the number of timeline generations(the number on the right) and get c

so called the combination

u use permutation when choice order(timeline order) matters

[rare fiber]

There’s a couple more corrections to make. you can’t make all these cases, and you’ve got to complete your process. I don’t think You can ignore the ways of choosing the BMWs

[devout bison]

i don't need bmw possibilities am not asked for that

i sent the other bot explanation for accuracy

[rare fiber]

For example the 1A , 1R case is not possible as you need 6 BMWs(we only have 4) to choose 8 choose

[devout bison]

who said that

i don't remember anything like that in my explain

[willow bear]

chatGPT output should not be trusted.

ever.

zero exceptions. if you can verify it without chatGPT then it means you never needed chatGPT in the first place.

[devout bison]

well i sent it only cuz i tought my explain was mindmixing

[mossy quiver]

chatgpt doesnt know what the hell it is talking about

[devout bison]

also the one in file is ai generated

[rare fiber]

You did this case, then you asked them to replace permutation by combination, by which I inferred that you meant do 5C1 * 6C1? Did I misinterpret you?

[mossy quiver]

it is a language model. it is designed to create coherent sentences, it is not your google

[devout bison]

the writings upwards are mine

[willow bear]

your explanation was very "mind-mixing", but chatgpt would not help that at all.

[devout bison]

yeah which would make no difference at al

since 1!=1

what's the problem there

also i counted several mistakes in my own

so i simply gave the idea of the way of possibilities

anyways

@static mesa u can ask again

there are more people

more miinds

oh damn' man

am gonna delete chat gpt

what the hell is this

[willow bear]

word of advice: if you are not 120% confident that you know how to solve a problem, then it's better not to try to help at all.

[devout bison]

ik how to but i may have done calculus or mind mistake

as i said earlier

[willow bear]

this metaphor will be unpleasant but it needs to be said:

what you have taken is basically the textual equivalent of a huge pile of shit, and we now have to live with it and smell it all.

[devout bison]

my mission was to give the idea of doing it

[willow bear]

and somehow shovel it all away.

[devout bison]

instead of talking maybe u help them

also u can't say the idea was wrong

so ig he kinda understood the method and

here we wait for u

[willow bear]

there may be a grain of truth there but it's buried under a pile of shit

[devout bison]

what is ur explain

[willow bear]

this is the original problem:

jamies dad allows him to choose 8 cars for his birthday 

jamie's dad has 6 aventadors , 5 r8s , 4 bmws 

in how many ways can jamie pick such that there're an equal ammount of r8s and aventadors

[devout bison]

yeah

[willow bear]

we are choosing 8 cars from this set with the condition that the number of R8's and aventadors

if we choose a aventadors, b r8's and c BMW's, the number of choices will be C(6, a) * C(5, b) * C(4, c). it remains to be seen what values of a, b and c are possible.

we know that a=b and that a+b+c=8, leaving us with the following possibilities:

• (0,0,8) -- rejected: not enough BMWs
• (1,1,6) -- rejected: not enough BMWs
• (2,2,4) -- possible
• (3,3,2) -- possible
• (4,4,0) -- possible

[devout bison]

so u done?

[daring tapir]

(Not that relevant to the problem but doing it as if all the cars are unique would be a fun exercise)

[devout bison]

thats exactly what we do. we look onto each as unique

[willow bear]

i stopped one step short of the solution.

[daring tapir]

Nope like in bmws they are numbered bmw 1 bmw 2... bmw4

[willow bear]

if OP cares about solving the problem and understanding it, they will complete the last step

if not, they will not

[devout bison]

c'mon seemingly u couldn't huh

[willow bear]

???

[devout bison]

thats what i said bruh

[daring tapir]

Rude ppl argh

[daring tapir]

Thats not what we are doing rn though?

[devout bison]

not with u

[daring tapir]

We are considering all bmws as one, all r8s as one and all aventadors as one

[devout bison]

she has that behavior that plagues me also she didnt help just blab

[daring tapir]

And thats what we should do considering this problem statement

[devout bison]

@willow bear we are waiting for u

especially @static mesa

[willow bear]

waiting for what?

for me to give the answer?

[daring tapir]

Bro shes so helpful seriously

[devout bison]

nooooooooo

[devout bison]

exactly

[willow bear]

i gave you everything needed to get the answer yourself

the fact you're choosing to pester me for the last step is entitlement on your part

[daring tapir]

Exactlt

[daring tapir]

Yes we are?

[willow bear]

you know the saying "you can take a horse to water but you cannot make it drink"

[onyx panther]

You guys are getting out of control.

[daring tapir]

Also doesnt this belong in #competition-math anyways

[devout bison]

a grain of truth in

[devout bison]

a pile of blabbing

this is not mine but

@static mesa is probably on exam

he said he only needed combinations

[daring tapir]

Th? Why are we even helping them in that case?

[devout bison]

thats why we are waiting @willow bear

@willow bear

@willow bear

@willow bear

[daring tapir]

Someones begging to be banned

[willow bear]

<@&268886789983436800> we got some kinda harassment going on here

pingspam?

[devout bison]

what is a ping spam

[willow bear]

what you just did

[coarse marsh]

@devout bison please stop repeatedly spamming and pinging Ann

[devout bison]

also u typed on ur own upwards that u could be pinged

[devout bison]

okay

she said that she could be pinged

anyways

<@&286206848099549185>

[coarse marsh]

No one needs to be pinged 4 times in a row

[devout bison]

may be instead of blabbing <@&286206848099549185> help them

since this is a math server

not blabbing course

[daring tapir]

They got it already bruh

[devout bison]

still that is a math site not blabbing

[daring tapir]

Soo?

[coarse marsh]

Ann has laid out a substantial amount of help for the OP. We don't just straight up give out answers here so idk what you're waiting for, you are currently the one blabbing

[devout bison]

by just typing down what has given? and even so, what's the case with ann anyways

[coarse marsh]

Stop

[devout bison]

i pinged her for the answer u said don't and that's all but what's the case of urs anyways

[willow bear]

i did not "just type what was given".

[devout bison]

anyways thats none of the stuff but ig someone needs help in #help-44｜stanley-🌲-v2-dans

[hallow lodge]

does that apply to trinomials too

[fair basin]

Can I ask a high school calc question here?

[hushed sphinx]

Use #calculus for calculus.

[narrow oxide]

I don't need help here but I enjoyed this problem

[silent oasis]

Why did he use less than equal instead of less than sign?

[willow bear]

strict inequality is not a concern here

sure you could say these inequalities are strict. but (a) nobody gives a shit and (b) there is no obligation to write a statement that is as strong as humanly possible

[silent oasis]

So, how did he reach third line from the second one?

[willow bear]

take reciprocal of everything

[silent oasis]

does it change the sign?

[willow bear]

it flips the sign

[silent oasis]

from less than to greater than

do you have any example

[willow bear]

example of what?

[silent oasis]

how the flipping happens

[willow bear]

are you asking why 0 < a <= b implies 1/a >= 1/b...?

[silent oasis]

yeah

[willow bear]

like the formal proof?

[silent oasis]

yes.

[willow bear]

a <= b implies b - a >= 0

1/a - 1/b = (b-a)/(ab) = (b-a) * 1/a * 1/b >= 0

b-a >= 0, while 1/a and 1/b are each positive
thus 1/a - 1/b >= 0

is this what you wanted

[silent oasis]

yes

[silent oasis]

i didn't understand what you did here in this line

are 1/a and 1/b positive because both a and b are greater than zero?

[willow bear]

yes

[hushed sphinx]

The conclusion isn't even true if a and b have different signs.

[viscid thistle]

hey, sorry to interrupt but if i have the quadratic approximation (second order taylor expansion) of cos(x)

how do i find the quadratic approximation of cos^(1000) x?

approximately

[hushed sphinx]

If a and b are both negative, the conclusion is true again, but we'd need some modifications to the reasoning Ann showed.

[hushed sphinx]

Binomial theorem, I'd say.

[viscid thistle]

i thought of that but it didn't work out

cos(x) = 1- x^2 + x^4/4! + ... right?

(1 - x^2 + x^4/4! + ...)^1000 = ?

it's a bit hard

[hushed sphinx]

Or perhaps just wing it:
If you imagine multiplying out (1 - x^2/2 + o(x^2))^1000, there will be some times that contain one or more factors of o(x^2). You're not interested in them; they'll have too high degree to matter for your quadratic approximation. Then there will be terms containing factors of 1 and factors of -x^2/2. There you're interested in exactly those that have one factor of -x^2/2 and 999 factors of 1. How many of those are there?

[viscid thistle]

yeah i think i'm aware of it but i'm probably missing something basic of what i should be doing

like can we try with sin(x) for example?

what am i looking to do exactly

[hushed sphinx]

I don't think I have any better explanation ready than what I just said.

[viscid thistle]

no it's not about your explanation, but i think it just didn't "click" for me

I agree with the extra O(x^2) term part

every terms of higher degree gives you an even higher degree so we don't care

But the last sentence i'm not sure how to put to practice

[hushed sphinx]

So you agree that the quadratic approximation you're looking for will also be the quadratic approximation of (1-½x²)^1000?

[viscid thistle]

whta about the O(x^2)? do we still not keep it there just so that after our expansio we have something ike O(x^1004)?

which we don't care for but yeah, do we still not leave it there regardless?

[hushed sphinx]

Oh, I thought those were the ones you jsut agreed could be ignored.

[viscid thistle]

Yes well technically i can ignore, i just thought formally you'd write O(x^2) as well right?

[viscid thistle]

but yes i agree

[viscid thistle]

so just 1 - 1000 (1) ( x^2) + ...?

wait so just x^2

wait i'm being stupid probably lol

okay nvm

[hushed sphinx]

Okay, so let's write that out with the binomial theorem:
$$ (1 + \tfrac12 x^2)^{1000} = \sum_{k=0}^{1000} \binom{n}{k} 1^{n-k} \left( \frac{x^{2k}}{2^k}\right) $$

[obsidian monolith]

Troposphere

[hushed sphinx]

The k=0 term is just 1.

Every term with k>=2 has too high degree to care about.

[viscid thistle]

true

okay i think this part is fine

what about more termst hough

let me think of a question

[hushed sphinx]

So there's now only one term left, namely the k=1 term.

[viscid thistle]

okay so what about for sin(x)

this time it's a cubic approximation for anything meaningful

because it has no quadratic term

[hushed sphinx]

Yes, this is right. (Sorry, didn't see it while I was typing).

[viscid thistle]

well shouldn't the second coeffficient be like 500?

[hushed sphinx]

Ah, right. Sorry.

[viscid thistle]

anyway no problem i get the idea i think my main problem is "more terms" in the expansion

by more terms i mean

multinomial expansion i guess not binomial anymore

[viscid thistle]

this is why i was curious how we'd do it with say O(x^3) formally put into this

[hushed sphinx]

Do you know multinomia expansion (which would make it easier) or were you just guessing the word?

[viscid thistle]

cos(x) = 1 - x^2/2 + O(x^3)

i'm curious what would be the right way to write out: (cos(x))^1000 = 1 - 500(x^2) + O(x^3)

[viscid thistle]

i've heard of it lol

but not too familiar

[hushed sphinx]

That looks right.

[viscid thistle]

but like if i were to reason this out

it'd be: (1 - x^2/2 + O(x^3))^ (1000) right?

how do you deal with that?

[hushed sphinx]

I'm confused. Wasn't that was we just did?

Or you're asking for a formal justification for ignoring the O(x³) terms?

[viscid thistle]

i know but our argument was that we can just ignore O(x^3)

which is sound yes

but like what if you just want to incorporate that into our expansion

instead of outright deleting it

so instead of writing $\qty ( 1 - \frac{x^2}{2})^{1000}$, how would we do it with the presence of $\qty ( 1 - \frac{x^2}{2} + o(x^3))^{1000}$

[obsidian monolith]

!Kiz__

[hushed sphinx]

Hmm, let's define f(x) = 1 - x^2/2 for brevity.
Then we're looking at (f(x) + O(x^3)) ^ 1000, and we want to argue that this is f(x)^1000 + O(x^3).

Binomial theorem again: $$ (f(x) + O(x^3))^{1000} = \sum_{k=0}^{1000} \binom{1000}{k} f(x)^{1000-k} O(x^3)^k $$

[viscid thistle]

oh i see, so we can always just turn it into a binomial

[obsidian monolith]

Troposphere

[hushed sphinx]

Indeed.

[viscid thistle]

okay nice thanks, but wait my actual context is to use taylor on limits

therefore i know something like sin(x) approximates to x - x^3/6 + o(x^5) right?

[hushed sphinx]

O(x^5) not o(x^5)

[viscid thistle]

would small o notation not be the same as big O notation here?

[hushed sphinx]

Or alternatively ... + o(x^4).

[viscid thistle]

oh

[hushed sphinx]

No -- for example, x^5 is O(x^5) but is not o(x^5).

[viscid thistle]

okay then i don't know what little o notation is

can u briefly explain it then?

why o(x^5) isn't x^5?

[hushed sphinx]

We say that f(x) is o(g(x)) if f(x)/g(x) goes to 0 as x->0.
We say that f(x) is O(g(x)) if f(x)/g(x) is bounded near x=0.

In other words, o(x^5) stands for a function that goes to 0 strictly faster than x^5 does.

But O(x^5) means one what goes to 0 at least as fast as x^5 does.

[viscid thistle]

Oh i see, i get it

nice

okay so i think back to my question

sin(x) = x - x^3/6 + o(x^4)

how does this follow:

that's from stackexchange (https://math.stackexchange.com/questions/1786112/how-to-calculate-lim-x-to-0-fracx6000-sin-x6000x2-sin-x)

[willow bear]

see #calculus

i just explained this exact same q to another person

[viscid thistle]

oh great, thanks!

[silent oasis]

what modifications?

[hushed sphinx]

I'm too lazy to go through and figure out exactly where they are, but I imagine there might be some signs that need adjusting.
And instead of "because 1/x and 1/y are both positive" we'd need to say "because 1/x and 1/y are both negative, their product is positive, ad therefore ..."
Not deep changes, just pesky.

[willow bear]

well they are deep if you have zero familiarity with proofs or zero familiarity with inequality manipulations

[hushed sphinx]

Fair.

[daring tapir]

Wouldnt anyone trying precalc have a good enough knowledge of inequalities anyways?

[agile sparrow]

not me

[daring tapir]

Thats a fucking lie lmao

[agile sparrow]

I still crossmutiply with problems of < and >

Aand cancel terms each side

[daring tapir]

We were taught inequalitirs the first day we started jee prep

[agile sparrow]

I'm not him

I'm himm

[daring tapir]

Ahem is there smthn else to do?-

Not considering like the triangle inequality or am-gm shit

[agile sparrow]

u can't do that

[daring tapir]

We can if we take care of the sign

At least afaik

[agile sparrow]

(x-2)/(x-4) > (x-6)/(x-4)

then u have to take any one to the other side

U can't just cancel out the denominator

[daring tapir]

You could just cancel? As long as x-4 is positive anyways

If its not you can still cancel

Just change the sign

Ig

[agile sparrow]

U can't

[daring tapir]

Uh?

[agile sparrow]

nuh uh

[daring tapir]

Lmao okay

[agile sparrow]

nuh uh

[naive marsh]

Anyone have a good strategy for review the AP test

[narrow briar]

hi! im on highschool and i love maths. i like functions but the content on the highschool lessons seems very easy and i want to learn more about calculus. I learnt limits and derivatives by the textbook by my own, but the book doesnt explain integrals. Can someone tell me how can i learn integrals and that kind of things? thank you

[daring tapir]

Thomas/stewart etc etc are all good books since you can learn from books on your own

[hushed sphinx]

Plan A would be to ask your teacher if they can recommend some material. Schools sometimes have old advanced textbooks sitting on a dusty shelf somewhere.

[narrow briar]

thank you

and do you know any web or youtube channel for it?

[hushed sphinx]

If you're working that far ahead, you might consider if you should be looking at some proof-based real analysis instead of just forging ahead to integrals.

[daring tapir]

Proff leonard is a good one
I see mathisfun.com being recommended to ppl a lot too but havent used it personally so cant comment on its quality

[narrow briar]

what is proof-based real analysis?

[hushed sphinx]

Limits, derivatives, etc, but presented with theorems and rigorous proofs of everything instead of "here's the steps you must follow".
For all I know, that might actually be what you've already read -- it varies a lot between parts of the world whether one even distinguishes between "calculus" and "real analysis" in this way.

[narrow briar]

that seems nice

[hushed sphinx]

As a quick touchstone example, do you know what the intermediate value theorem says -- and would you have an idea of what would be necessary to start proving it?

[narrow briar]

oh, i dont know it

[hushed sphinx]

Good, that means what you've been learning is not proof-based real analysis yet, and my suggestion to look into that won't waste your time. :-)

This is a topic where you'll pretty much need to use books rather than videos as your learning materials, though.

If you're on good terms with your teacher (and the teacher is somewhat competent), I'll repeat my suggestion to ask them. The teacher knows you better than we do, and can make a more educated guess at whether you should just be thrown straight at Baby Rudin or find a softer-and-friendlier introduction instead.

If that doesn't pan out, ask for concrete recommendations in #calculus or #book-recommendations.

[urban stream]

using this bc I couldn't find a channel dedicated to geometric series

is this correct?

its the evaluation of this sum

[hushed sphinx]

!nogpt

[tender quest]

Please do not trust ChatGPT or similar AI tools for mathematical tasks, as they often generate output which "sounds correct" but has numerous factual or logical errors. Use of these AI tools to answer other people's help questions is strictly against server rules (see #rules).

[hushed sphinx]

In this case I don't immediately see any error, though.

[urban stream]

yeah it was incorrect

i got the correct one now haha

i think this is the one

$S = \frac{15}{2} \left(1 - \left(\frac{3}{5}\right)^n\right)$

[obsidian monolith]

uh

[hushed sphinx]

If I set n=1 there I get S=3, which does not seem to match what you said you're evaluating.

[grim elk]

How hehe

[urban stream]

huh

shit I'm gonna have to do it again now

[urban stream]

anyone any idea how i'd go about evaluating this

[willow bear]

are you 110% sure that you wrote it correctly and made zero typos?

@urban stream

asking because the probably-intended meaning of the sum doesn't align with the sum as it is written right now: $$\sum_{r=1}^n \frac{15}{4} \cdot \paren{\frac{3}{5}}^n$$

[obsidian monolith]

|Ann⟩

[hushed sphinx]

(What Ann is saying is that it looks extremely likely that the n'th power in that formula should have been an r'th power instead).

[urban stream]

no i think its incorrect

[urban stream]

OH

FUCK

[willow bear]

ok so then can you post the correct sum

[urban stream]

the sum of the terms are \sum_{r=1}^{n}\frac{15}{4}\left(\frac{3}{5}\right)^{r}-1

$\sum_{r=1}^{n}\frac{15}{4}\left(\frac{3}{5}\right)^{r}-1$

the -1 is gonna be with the r, couldn't type it on with latex

$\sum_{r=1}^{n}\frac{15}{4}\left(\frac{3}{5}\right)^{r-1}$

[obsidian monolith]

uh

[urban stream]

got it

@willow bear

[willow bear]

ok

so

!status

[tender quest]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[urban stream]

2

the question later on says to find the least value of n for which sum of infinity - sum of nth terms <0.045

[bold wren]

Hey people someone told me to come to this channel maybe i will find what i want but promise you will not laugh after i say my request? I really need it but dont ask why, ok?

[urban stream]

no need to be ashamed

[bold wren]

Is there any function that has like one point like for example (1;3) and that's it no more points?

[hushed sphinx]

Many qustions cannot be answered well without knowing why, though.

[bold wren]

I just need it

[upbeat wyvern]

How do I calculate the area of this pool using calculus?

[urban stream]

this goes in #calculus

[upbeat wyvern]

Aight

[urban stream]

This was the whole q

,rccw

[urban stream]

Struggling with C

[winter comet]

you could pretend its 2 curves, put it on a coordinate axis, and use riemann sum to guess

you really don't have any lengths or units tho

soo

imposibleh

XD

[golden cairn]

How do I solve this?

[hushed sphinx]

Didn't you already ask that yesterday?

And if you keep asking the same question, people will just keep giving you the same answer that you didn't find helpful the previous times.

Would you be able to solve this equation: 3·x·1.792 = (2x-3)·0.693?

[golden cairn]

Yeah

I got -0.52

That’s the same answer for the question I asked

So you got me the correct answer but I don’t know what you did

[hushed sphinx]

The previous times you asked, you got to
3x·log(6) = (2x-3)·log(2)
and then got stuck.

[golden cairn]

Yeah

[hushed sphinx]

I simply calculated the values of the logarithms: log(6) = 1.792 and log(2) = 0.693.

(Though now I see I should probably have used the base-10 logarithm instead of the natural one, sorry).

[golden cairn]

Yeah but the teacher said we can’t use calculators

Is this the next step?

[hushed sphinx]

Hmm, without a calculator, I don't think there's any reasonable way to reach the value -0.52 as an answer.
(I'm assuming you don't have a slide rule or a log table either ...)

[golden cairn]

Nope

[hushed sphinx]

No, you're misapplying some log rules there. The next step would be to rearrange as
(3·log6) x = (2·log2)x - 3·log2

[winter comet]

they prob want you to distribute and solve for the exact value of x

[golden cairn]

Yeah I already did that

But I skipped a step

And just showed what I thought might have been the next step

[winter comet]

its not

[golden cairn]

I know I thought so

[winter comet]

2xlog2 - 3log2 isn't that

[golden cairn]

I tried using these

[winter comet]

you should subtract 2xlog2 from both sides instead

[golden cairn]

Okay

[winter comet]

that only works if its M or N inside the log

[knotty pike]

use log6(6)³^x in both sides

[winter comet]

2x and 3 are both outside the log function

[winter comet]

no wait why

[hushed sphinx]

I imagine when you solved my version with decimal numbers, your next step was to combine all the terms with x in a single term.

[winter comet]

oh nah that step was wrong in the first place

[golden cairn]

Yeah

[winter comet]

then factor out x

and then divide by everything thats left

to solve for x

[hushed sphinx]

So the point here is that log(6) and log(2) are just two particular numbers. They behave the same in an equation as 42 and 18827 would have done.

[golden cairn]

I am trying that

Right now

[knotty pike]

it's correct
3x = 2log6(2)x - 3log6(2)

[hushed sphinx]

So what you should be thinking first when you see

(3·log6) x = (2·log2)x - 3·log2
is not "uh oh there's a logarithm there, I need find and use a log rule", but instead "ho hum, there's some constant number there, it doesn't affect how I solve the equation precisely what the constant is".

[winter comet]

thats correct...

[golden cairn]

Okay

[winter comet]

idk what this is tho

[hushed sphinx]

It looks pretty incorrect to me. For example, if you plug in x=0 it claims that 0 = -3log(6)·2.

[golden cairn]

I got the answer

I did the factoring

[winter comet]

oh ur right its not nvm 🗿

[winter comet]

what u get

[golden cairn]

I just factored out the x and then did (-3log2)/(3log6-2log2) and then I got -0.5213

[hushed sphinx]

Perfect.

[winter comet]

yeh

[golden cairn]

Thanks

[knotty pike]

[3-2log6(2)]x = -3log6(2)

x = - (3log6(2)/3-2log6(2))

x = - (3log6(2)/log6(6³)-2log6(2))

x=-(3log6(2)/log6(6^3.2^-3))

doing that u got:

x=-(3log6(2)/log6(27.2))

x=-(3log6(2)/log6(54)

x=-3log54(2)

x=-0,5213

[golden cairn]

How do I solve this?

I tried the same method

But it won’t work this time

I keep getting the incorrect answer

[hushed sphinx]

The same general approach should work. Can you walk us through your steps that give an incorrect answer?

[golden cairn]

Okay so I got up to this step: xlog(3/2)+xlog5=2log5

I factored out the x

[hushed sphinx]

Smaller steps, please?