help-4

[night patio]

each 3-length group is permuted in 3! = 6 ways in these 60 ways

we can divide 60 by 3!

or 60/6 = 10

[misty vessel]

Ohh ok

[night patio]

yes now what if we make 4-length permutations

how many can you make?

[misty vessel]

4!= 12

[night patio]

with 5 digits

how many can you make

[misty vessel]

120

[night patio]

also we can say (5!/2!)/3! right?

[misty vessel]

Yes

[night patio]

yes

Group 1 (A, B, C, D):
ABCD, ABDC, ACBD, ACDB, ADBC, ADCB,
BACD, BADC, BCAD, BCDA, BDAC, BDCA,
CABD, CADB, CBAD, CBDA, CDAB, CDBA,
DABC, DACB, DBAC, DBCA, DCAB, DCBA

Group 2 (A, B, C, E):
ABCE, ABEC, ACBE, ACEB, AEBC, AECB,
BACE, BAEC, BCAE, BCEA, BEAC, BECA,
CABE, CAEB, CBAE, CBEA, CEAB, CEBA,
EABC, EACB, EBAC, EBCA, ECAB, ECBA

Group 3 (A, B, D, E):
ABDE, ABED, ADBE, ADEB, AEBD, AEDB,
BADE, BAED, BDAE, BDEA, BEAD, BEDA,
DABE, DAEB, DBAE, DBEA, DEAB, DEBA,
EABD, EADB, EBAD, EBDA, EDAB, EDBA

Group 4 (A, C, D, E):
ACDE, ACED, ADCE, ADEC, AECD, AEDC,
CADE, CAED, CDAC, CDEA, CEAD, CEDA,
DACE, DAEC, DCAE, DCEA, DEAC, DECA,
EACD, EADC, ECAD, ECDA, EDAC, EDCA

Group 5 (B, C, D, E):
BCDE, BCED, BDCE, BDEC, BECD, BEDC,
CBDE, CBED, CDBE, CDEB, CEBD, CEDB,
DBCE, DBEC, DCBE, DCEB, DEBC, DECB,
EBCD, EBDC, ECBD, ECDB, EDBC, EDCB

these 120 right?

[misty vessel]

Yes I suppose

[night patio]

now see each 4-length permutation is permuted in 4! ways

we can divide 120 by 4!

[misty vessel]

Wait wait

[night patio]

to get the number of groups

[misty vessel]

Can you elaborate this

[night patio]

yes can you give me 5 mins, I was busy with something important

sorry

[misty vessel]

Yeah yeah

No prob

[night patio]

@misty vessel

hi

[vale dock]

@misty vessel Has your question been resolved?

[marble bronze]

its the same as 5C4 or a combination right?

[night patio]

yes

[marble bronze]

oo okayy thanks

[night patio]

I was trying to show him how to get the combinations result

[marble bronze]

owhhh

[night patio]

@misty vessel

[misty vessel]

Yes

Sry sey

Sry

Can we continue

My bad for showing up late

@night patio you there?

[night patio]

yes hi

[night patio]

Okay we have these 120 permutations

in case of 4-digit or 4-length permutations using the 5 digits ABCDE

[misty vessel]

Ok

[night patio]

you can make 120 4-length permutations from these 5 digits using the formula 5!/4!

right?

[misty vessel]

Yes

Right

[night patio]

now can you see like when you do 5 * 4 * 3 * 2

[night patio]

you make these pairs

[misty vessel]

Yes ok

[night patio]

and we can arrange the pairs in this way

these groups

each group has combinations

which are permuatations of each other

[misty vessel]

Ok

[night patio]

like group 1 contains the combination of ABCD

[misty vessel]

Yes

[night patio]

group 2 contains ABCE

[misty vessel]

Yes

[night patio]

now i know that we are making 4-length permuations

so i will get 4! permutations for each combination

right?

[misty vessel]

Yes

Right

[night patio]

so if each group has 4! = 24 permutations

and I know that total permutations which is 5!/4! = 120

[misty vessel]

Yes

[night patio]

I can divide 120/24

to get the number of groups

[misty vessel]

Ohhh

Okk ok

[night patio]

now look at the other cases before

[misty vessel]

I gt it

[night patio]

in the case where we made 3-length permutations

Group 1 (A, B, C):
ABC, ACB, BAC, BCA, CAB, CBA

Group 2 (A, B, D):
ABD, ADB, BAD, BDA, DAB, DBA

Group 3 (A, B, E):
ABE, AEB, BAE, BEA, EAB, EBA

Group 4 (A, C, D):
ACD, ADC, CAD, CDA, DAC, DCA

Group 5 (A, C, E):
ACE, AEC, CAE, CEA, EAC, ECA

Group 6 (A, D, E):
ADE, AED, DAE, DEA, EAD, EDA

Group 7 (B, C, D):
BCD, BDC, CBD, CDB, DBC, DCB

Group 8 (B, C, E):
BCE, BEC, CBE, CEB, EBC, ECB

Group 9 (B, D, E):
BDE, BED, DBE, DEB, EBD, EDB

Group 10 (C, D, E):
CDE, CED, DCE, DEC, ECD, EDC

we had this

[misty vessel]

Yes

[night patio]

each group has 3! permutations

why?

[misty vessel]

Because that's the maximum number of possible arrangements of them

[night patio]

yes we have 3 digits

so 3! ways to permute them

[misty vessel]

Yea

[night patio]

and I know there are a total of 5!/2! = 5 * 4 * 3 = 60 3-length permutations

[misty vessel]

Ok

[night patio]

so I divide 60 by 3!

because i know that each group would have got permuted in 3! ways when I did 5 * 4 * 3

so 60/6 = 10

[misty vessel]

Yes right

[night patio]

we get 10 groups

[misty vessel]

Ok

Ok

[night patio]

and you can see we have 10 groups

[misty vessel]

Yeah I get it now

[night patio]

Now we can say

we just found a method

to make k-length combinations from n distinct objects

like you now know how to calculate the number of 3-length combinations from 5 distinct objects right

[misty vessel]

Right

[vale dock]

Channel closed

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[night patio]

just make total 3-length permuatations, that would be 5!/(5-3)!

.reopen

[misty vessel]

Ohh

So this is the formula for it?

[night patio]

small correction

the formula you said earlier

n!/(k-1)!

it should be

n!/(n-k)!

for k-length permutations

[vale dock]

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[night patio]

@misty vessel do .reopen

[misty vessel]

Ok

.reopen

[night patio]

wait this is my help channel now

nvm

np

yea go on

[misty vessel]

Wha 😂

[night patio]

where were we

[misty vessel]

Here

He formula

The*

[misty vessel]

Ohh ok

[night patio]

yes you know what n!/(n-k)! gives us right

[misty vessel]

Ok

I'll update it

[night patio]

like if i want to know number of 3-length perms for 5 digits

I will get 5 * 4 * 3

or 5!/2!

[misty vessel]

Yes

[night patio]

n!/(n-k)!

[misty vessel]

Ok

[night patio]

this is general formula for n objects and k length

[misty vessel]

So that's how we got 2

Ok

[night patio]

yes

[misty vessel]

Hey wait

[night patio]

now notice

[night patio]

yes

[misty vessel]

I am trying to understand but I am really sleepy and when I am sleepy, I am sleepy

[night patio]

all good

you can sleep hahaha

[misty vessel]

You can tell me this if you want

Yeah

[night patio]

yes last thing

[misty vessel]

😂

[night patio]

out of these n!/(n-k)! permutations

each group would have been permuted k! times

[misty vessel]

Hmm

Ok

[night patio]

like this

[misty vessel]

Ah ok

[night patio]

each 3-length combination is permuted 3! times

so each k-length combination would have permuted k! times

and we can group them together

[misty vessel]

Ohh ok

[night patio]

we have n!/(n-k)! total perms

[misty vessel]

Ohh okaye

[night patio]

so we can divide n!/(n-k)! by k!

we get this

[misty vessel]

Yes

[night patio]

this will give you the total number of combinations

[misty vessel]

Ohhh

[night patio]

like if i tell you there are 10 students in a class

How many groups of 2 can you make?

thats just 10!/(2! 8!)

[misty vessel]

Yes

[night patio]

these are the total number of combinations

do you get what is the difference between permutations and combinations now?

[misty vessel]

Ok

I do get it

[night patio]

this is usually known as n choose k

what we did is we derived it

[misty vessel]

Ok

Yeah

[night patio]

you can skip the derivation if you want

but i like it because it gives a lot of intuition

[feral roost]

hi

[misty vessel]

Is it hard ?

To derive it?

[night patio]

we just derived it right?

[misty vessel]

Like with all the sentences?

[night patio]

this formula

[misty vessel]

.

[night patio]

no we know that

[misty vessel]

Ok

[night patio]

there are n!/(n-k)! total k-length permutations for n distinct objects

[night patio]

do you understand this part?

[misty vessel]

Yes

Absolutely

[night patio]

now each k-length permutation can be permuted in k! ways

[misty vessel]

Yeah

[night patio]

so we have k! duplicates or permutations of each k-length sequence

[misty vessel]

Ok

Man please

Can we close 🙏🏻🙏🏻🙏🏻

[night patio]

yea

😭 😭

we should

[misty vessel]

😭😭

Yeah

[night patio]

sorry for complicating

goodnight

😭

[misty vessel]

Nah nah

That's fine

I like it

You really helped me man

Thanks

[night patio]

all good hahaha

[misty vessel]

: )

Bye

.close

[night patio]

.close

[vale dock]

Channel closed

Closed by @night patio

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[night patio]

hi

[surreal relic]

umm sorry

i though u r the bot then noticed the Math tag

[night patio]

😭 oh

actually, i am the bot

[surreal relic]

hehe 😅 why do u say so?

[night patio]

jk 😭

[vale dock]

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[midnight pier]

I am not sure how to get ax^2 and bx, i know that c = -1

[woeful trench]

You could e.g. get a set of simultaneous equations by the fact that -1 and 4 are roots

[midnight pier]

ah yea i tried that but i kept getting stuck cause they = 0

im kinda rusty on simulatanious eqs

[woeful trench]

They being the output you find?

[midnight pier]

well if its a root then f(x) = 0

i got confused doing simultanious eqs

[woeful trench]

Did you rearrange the equations you found after you substituted -1 and 4 into those, and setting the output equal to zero? You could get them closer to the "usual" form that simultaneous equations come in, and that may make it better(?)

[midnight pier]

i found the solution online too

i mean i did some of it

just wasnt sure on last part

[hearty belfry]

the reason why there is a 1/4 is because we had to transform x^2 - 3x - 4

a stretch parallel to the y axis by scale factor of 1/4

so that the -4 (this is the y intercept for x^2 - 3x - 4) can turn into -1 (our c which is the y intercept we want)

[midnight pier]

yes thank you

[vale dock]

@midnight pier Has your question been resolved?

[vale dock]

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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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• Be polite and have a nice day!

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[ruby oriole]

How do I do question 19 with ncr and is ncr an optimal way to do this if my concern is speed?

[hardy coral]

What's ncr?

[sharp whale]

nCr is way to type the binomial coefficient (n choose r)

[ruby oriole]

n!/r!(n-r)!

Idek what npr is and what it does tbh

Might learn it later

But is ncr efficient for this question or does it make it harder?

Tbh I kinda forgot this slightly

[sharp whale]

nPr is for permutations and is n!/(n-r)!

[bold orchid]

Does anyone know a good algorithm for checking if there are roots in a specific interval of a non-linear function?

[ruby oriole]

Ts not an available channel 🙏

[sharp whale]

theres a lot of reading on that you can find online, you can start with newton's method

[ruby oriole]

Newton Raphson thing

[hardy coral]

Idk, to me the most straightforward way is just to enumerate the ways to make 5, 6, 7, and divide that by the total possible rolls

[sharp whale]

you should open a new channel (read #❓how-to-get-help) though

[hardy coral]

Might be nCr in disguise but it's really easy here

[ruby oriole]

I need to know how to do this sort of thing in under 90 secs in about 58 days

[ruby oriole]

Is this one of the harder cases of ncr

Cause I lowk have no idea how to do it with ncr

[sharp whale]

this is one of the cases where nCr is not necessary

[ruby oriole]

I only know how to do ncr for the cases like question 18

[hardy coral]

I don't know how to use nCr here that isn't just for trivial computations

[sharp whale]

oh btw hey nel

havent seen you in a while

[ruby oriole]

So I shouldn’t worry about it if I can’t figure out how to do it

?

With ncr

I just thought that might cut down the time for me

[sharp whale]

it wont when the method to do this is extremely quick

are you familiar with

2  3  4  5  6  7
3  4  5  6  7  8
4  5  6  7  8  9
5  6  7  8  9 10
6  7  8  9 10 11
7  8  9 10 11 12

[ruby oriole]

That’s one of those tables

Forgot the name

When you add it up

[hardy coral]

Probably just get used to visualizing things like that table

[sharp whale]

when you consider the possibilities of adding two dice up

you get this table

[ruby oriole]

Yh

[sharp whale]

(1, 2, 3, 4, 5, or 6) + (1, 2, 3, 4, 5, or 6)

[ruby oriole]

Oh so I literally count?

[sharp whale]

now you can see in the table the diagonals where 5, 6, or 7 lie

yea

[ruby oriole]

Fairs

[sharp whale]

so its (6 + 5 + 4) / 36

[ruby oriole]

How do I decide what’s n and what’s r in general

[hardy coral]

Very slightly different way to do it is to realize that you can make 5 with 1+4 and 2+3 and so on until you get to 4+1, which is obviously 4 ways; similarly you can make 6 in 5 ways, and 7 in 6 ways

No real need for the table but it's a nice visualization

[ruby oriole]

Thank you

But say in another general situation when ncr is required

What’s a method to figure out what I use for n and r?

[hardy coral]

I don't think there's really a method, you just need to think about it the right way

Visualizing helps a lot

Have you done questions 20 and 21?

[ruby oriole]

No

Aren’t they relatively simple probability questions?

Just feel the need to clarify I didn’t take 20 mins to do them 🤣

[hardy coral]

Well, yeah they are simple

[ruby oriole]

I might need to out in some ncr practise

And probs a lot of thing I need to review and find more efficient ways to do

27 questions in 40 mins I think

But these questions are from the simplest part

[hardy coral]

Wait why the doubling in both cases?

Oh nevermind

[ruby oriole]

I could’ve technically used ncr to figure out the 2 I multiply by if I wanted to complicate my life right?

[hardy coral]

Yeah

[ruby oriole]

Is it 2C1

I assume cause red and green so 2

But I don’t fully understand why it’s 2C1

Bruh why can’t I just do pure

What’s wrong with differential equations

50million distributions pmo bruh

What’s the negative binomial gonna be useful for

[hardy coral]

Well for question 20 you can just split the experience: have 2 identical bags of balls, pull one from the first and one from the second, but the order doesn't matter so you have twice as many ways to do it

[ruby oriole]

Oh right

It’s like how sets you have and how many elements you choose from each set?

[hardy coral]

No it's which set you consider first

If the two bags are labeled A and B, you have a choice between pulling from bag A first or from bag B first

[ruby oriole]

So there’s 2 options for that

Right

So wbt the r?

[hardy coral]

(that's if you want to pull the red ball first; equivalently you can have the choice of pulling the red ball first or the green ball first, doesn't matter from which bag)

[hardy coral]

In nCr? n would be 2 because two sets, r would be 1 because you're choosing one

[ruby oriole]

Wait if it’s the same bag and there’s replacement then surely there’s not an option?

[hardy coral]

Okay that might be more confusing than I wanted

If you consider just one bag, with replacement, then your choice is between picking a red ball first and picking a green ball first

[ruby oriole]

Wbt the blue ball first

That’s still a possibility right?

[hardy coral]

Yeah but that wouldn't satisfy the goal

[ruby oriole]

Oh it’s possible ways to do what we want

[hardy coral]

Yes

[ruby oriole]

So like we take from the bag twice so like 2

Acc nah

Wait

I’m confused again 😭

Did this too long ago

Is it 2 because 2 selections?

[hardy coral]

Just imagine doing it IRL; in scenarios where you've picked a red ball and a green ball, then it's either because you've picked a red ball first and a green ball second or because you've picked a green ball first and a red ball second

[ruby oriole]

Yes

Maybe I should think about this with a more complex situation

Maybe I’m getting confused cause it’s obvious there’s 2 ways to arrange 2 things

[hardy coral]

Yeah maybe

[ruby oriole]

I’ll figure it out by doing some practise

Thank you

[vale dock]

@ruby oriole Has your question been resolved?

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[gray patrol]

can someone help on this pls ?

[normal hollow]

What have u tried

[hearty belfry]

there's usually 3 ways to solve it

anyways yeah what have you tried

[gray patrol]

moving the -3 to the other side

[hearty belfry]

completing the square
quadratic formula

factoring

[normal hollow]

Wont help

[hearty belfry]

that doesn't really help

[gray patrol]

why ?

[normal hollow]

I prefer this over all

[hearty belfry]

yeah factoring is the quickest

[normal hollow]

Cuz it's quadratic and all ig(what's the right answer to this question)

Do u k middle term splitting? @gray patrol

[gray patrol]

completing the square is a good idea too

[normal hollow]

Yea

[hearty belfry]

the quadratic you've given can be factored though

[normal hollow]

Yea

[gray patrol]

so just by trying x = 1.5

[normal hollow]

And going through completing square can lengthen the process in this one

[gray patrol]

but is it ok to the doctor if i do it on that way ?

[hearty belfry]

completing the square or factoring?

[gray patrol]

completing the square

[normal hollow]

U mean?

[hearty belfry]

Probably, but you can do it quicker with just factoring

[gray patrol]

i was trying the numbers and i x = 1.5

[gray patrol]

here

[normal hollow]

But y

[hearty belfry]

do you know how to factor?

[gray patrol]

so it is ok if i, said to the doctor the answer for it it is just 1.5, and not steps included

[gray patrol]

no sorry

[normal hollow]

U dont need to apologize 🥲

[normal hollow]

Well u wont be given the answer usually

[hearty belfry]

there's also another answer

quadratics require two solutions

(if both the roots are real and it's not a repeated root)

[gray patrol]

so how can we do it by factoring

[normal hollow]

It can have 1 too if it's a perfect square(ryt?)

[hearty belfry]

yeah I forgot to mention that

[normal hollow]

https://youtu.be/2ZzuZvz33X0?si=xKPTsktFDG9vlkY5

Watch this if u can

[gray patrol]

i will thx alot <3

[normal hollow]

Np

[vale dock]

@gray patrol Has your question been resolved?

[ruby oriole]

The way I do it is multiply the first and last term

So -18

And find 2 numbers that multiply to this and add to the middle one

Once you find the 2 numbers

You get 2 brackets

(6x+a)(6x+b)

Then simplify the terms

So if 3x+6 then cancel it down to x+2

[vale dock]

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[slow oak]

hello

[slow oak]

can someome help me with this problem

i need heelp with dss

[untold sequoia]

find ABC

[slow oak]

how cn i do that

im conused

confused with this problem

[untold sequoia]

one sec

[slow oak]

alrrrrrrrrrr

[untold sequoia]

180*(n-2)

[slow oak]

alr

oh ye'

180

?

[untold sequoia]

no

[slow oak]

.

[untold sequoia]

n is the number of sides

[slow oak]

yea

doesnt abc have 3 sides

[untold sequoia]

abc is the angle u need

[slow oak]

yes

[untold sequoia]

formula i gave you is for regular ngon

internal angle

so n is 5

[slow oak]

yes ik

oh

oh i see

540

[untold sequoia]

and then u need to do something with it

divide by n

and you get angle abc

[slow oak]

alr

[untold sequoia]

180(n-2)/n

and then it's easy

[slow oak]

so there are 5 sides right

108?

[untold sequoia]

see what to do now?

[slow oak]

so abc is 108

what do i do next

[untold sequoia]

look at ∆ABC

[slow oak]

mhm

[untold sequoia]

ab and dc are sides of a pentagon

[slow oak]

yes

[untold sequoia]

(regular pentagon)

so

[slow oak]

what do i d

do

[untold sequoia]

what is the difference between a regular pentagon and irregular

[slow oak]

well

all sides and angles equal in a reglar

and in a irrgular

[untold sequoia]

right

[slow oak]

diffrent

[untold sequoia]

now look at AB and BC

[slow oak]

yes

[gray patrol]

thx <3

[slow oak]

they equal

?

[untold sequoia]

what can you tell about the ∆abc

[untold sequoia]

they are

[slow oak]

hm

is abc

216

[untold sequoia]

no

abc is 108

[slow oak]

oh

ok

[untold sequoia]

what kind of triangle is ∆abc

?

you know that ab = bc

sooo

[slow oak]

isoloces

?

[untold sequoia]

yes

so what can you tell about the angles

in that triangle

[slow oak]

they r equal to each oher

?\

[untold sequoia]

yes

stop asking if you know the answer

I'll tell you if you're wrong

[slow oak]

me>

?

[untold sequoia]

xd

[untold sequoia]

you can now find acd

[slow oak]

alr ty

.close

[vale dock]

Channel closed

Closed by @slow oak

Use .reopen if this was a mistake.

[slow oak]

i gtg

sorry

[vale dock]

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• Show your work, and if possible, explain where you are stuck.
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[mild cipher]

help

[vale dock]

No need to ask “Can I ask…?” or “Does anyone know about…?”—it’s faster for everyone if you just ask your question! See https://dontasktoask.com/

[tacit onyx]

[mild cipher]

gay

so i have a prob

solve the equation $$log_8 x\-4log_8 x = 2$$

[rocky lotus]

Communist Africa

[mild cipher]

help pls

@normal hollow @nimble crypt

[nimble crypt]

what do you get if you turn $2=\log_8$ ...fill in the blank in the log

[rocky lotus]

MichaelRafto

[nimble crypt]

i suggest you just subtract the 2 logs on the left, then turn 2 into a $\log_8$

[rocky lotus]

MichaelRafto

[mild cipher]

hm

[mild cipher]

huh ???????????????????\

[nimble crypt]

if you do these, then what do you get?

[mild cipher]

that's not the question

[nimble crypt]

yeah it's part of solving it

[mild cipher]

hm

say it again

[nimble crypt]

you want to solve $$log_8 x\ -4\log_8 x = 2$$, right?

[mild cipher]

2 = log_8

[rocky lotus]

MichaelRafto

[mild cipher]

na this $$log_8 x\-4log_8 x = 2$$

[nimble crypt]

huh alr

[rocky lotus]

Communist Africa

[nimble crypt]

that is what i wrote

[mild cipher]

oh

hello

[nimble crypt]

ye

[mild cipher]

I stopeed at

wait, i got confused

idk where to start

[nimble crypt]

first subtract the logs on the left

you can do that so why not

[mild cipher]

ok brb

im back

yea that $$log_8 x/x = 2$$

[rocky lotus]

Communist Africa

[nimble crypt]

?

[mild cipher]

yea

thats the question

i forgot abt 4

[hearty belfry]

using double dollar signs puts equations on their own line btw

[mild cipher]

what do i even do

[nimble crypt]

how did you get $log_8 x/x=2$?

[rocky lotus]

MichaelRafto

[mild cipher]

i forgot abt 4

thats y

[nimble crypt]

what do you have as of now?

[mild cipher]

just the question itself

[nimble crypt]

so we are on square one

[mild cipher]

yep

[nimble crypt]

what do you get if you subtract the logs on the left?

[mild cipher]

i'll get log_ x/x^4

[nimble crypt]

no i mean without using any log laws just subtraction

[mild cipher]

bruh

as how

[nimble crypt]

no need for log laws yet

[mild cipher]

hm

[nimble crypt]

you have $log_8 x - 4log_8 x$ right?

[rocky lotus]

MichaelRafto

[nimble crypt]

that

is just $-3log_8 x$

[rocky lotus]

MichaelRafto

[mild cipher]

ohn, due to the co-efficent

ooh

then it will be $$log_8 x ^-3 = log_8 2$$

[rocky lotus]

Communist Africa

[mild cipher]

and also when were done, i'll like to ask u something

[nimble crypt]

right side wrong man

[mild cipher]

????????????????????

[nimble crypt]

remember $1=log_8 8$

[rocky lotus]

MichaelRafto

[mild cipher]

yea

[nimble crypt]

if you have a smaller value than 8 in the log you get a result smaller than 1

[mild cipher]

huh

smaller

like

example

[nimble crypt]

$0.5=log_8 sqrt(8)$

[rocky lotus]

MichaelRafto

[mild cipher]

hm

alr

ur cooking

hm

[nimble crypt]

you have 2 instead of 0.5

so what's inside the log8?

[mild cipher]

explain it again

[nimble crypt]

$0.5=log_8 sqrt8$

[mild cipher]

because inorder to reduce a decimal to its lowest form we use -whole nu,ber

ok

half

ok

[nimble crypt]

(cmonnnnnn texit)

[mild cipher]

just like how $$1 = log_8 8$$

[rocky lotus]

MichaelRafto

Communist Africa

[nimble crypt]

ye

[mild cipher]

ok

wow

[nimble crypt]

so for 2 what do you have?

[mild cipher]

i have root 2

[nimble crypt]

for 0.5

[mild cipher]

or according to indices 1/2

wait, hter's no 0.5

[nimble crypt]

yes ik

we have 2

[mild cipher]

yea

$$log_8 x-4log_8 x = 2$$

[rocky lotus]

Communist Africa

[mild cipher]

thats the question

but im stcuk at the left hand side

[nimble crypt]

the left hand you just subtract

[mild cipher]

yea

but there a 4 behind the log

[nimble crypt]

it's like have x-4x

[mild cipher]

ok

according to log - = divide

[nimble crypt]

simple subtraction, no log laws required

[mild cipher]

oh

[nimble crypt]

no log laws required

[mild cipher]

because of the co-efficent

right

[nimble crypt]

so what do you have?

after you subtract

[mild cipher]

$$-3log_8 x$$

[rocky lotus]

Communist Africa

[mild cipher]

= 2

[nimble crypt]

yes

what do you do next?

[mild cipher]

then since 2 = log 2 2^2

[nimble crypt]

why did you take $log_2$?

[rocky lotus]

MichaelRafto

[mild cipher]

uhm

yea

so im right

[nimble crypt]

you can take $log_a$ with a being anything you want it to be

[rocky lotus]

MichaelRafto

[nimble crypt]

but in this exercise what do you want it to be?

[mild cipher]

8

[nimble crypt]

so you take $log_8$

[rocky lotus]

MichaelRafto

[nimble crypt]

now there is one part in the log8 that is missing

[mild cipher]

yea

the number

wait

the RHS is 2

and we wanna log it

so inorder to log an unlogged

we have to make th base exactly the same as the other or in the left hand side

right

[nimble crypt]

yes for $1=log_8$ tis $1=log_8 8^1$

[mild cipher]

alr

log ehn

wahala dey for u

[nimble crypt]

for $1=log_2 its $1=log_2 2^1$

[rocky lotus]

MichaelRafto

[mild cipher]

so $$2=log_8 8^2$$

[rocky lotus]

Communist Africa

[nimble crypt]

yes

ok so what do you have now and what do you do next?

[mild cipher]

since $$-3log_8 x = log_8 8^2$$

[rocky lotus]

Communist Africa

[mild cipher]

so im gonna cancel out the $$log _8$$

[rocky lotus]

Communist Africa

[nimble crypt]

ye

[mild cipher]

so its $$-3*x=8^2$$

[rocky lotus]

Communist Africa

[nimble crypt]

ye

no that is wrong, my bad man

[mild cipher]

yea

i figured

because it was too long

[nimble crypt]

yea you have to divide with -3 before you delog (and you don't have to turn 2 into a log)

[mild cipher]

ok

ohn

[nimble crypt]

after the subraction you just divide with -3

both sides

[mild cipher]

8 is not divisible by 3

[nimble crypt]

$-3log_8 x=2$

[rocky lotus]

MichaelRafto

[nimble crypt]

$log_8 x= -2/3$

[rocky lotus]

MichaelRafto

[mild cipher]

uhn

thats gonna be in decimal

wait

according to log its

gonna be

$x=8^-2/3$

[rocky lotus]

Communist Africa

[mild cipher]

that's 0.25

[nimble crypt]

yes

correct

[mild cipher]

my answer is 0.25

finally

[nimble crypt]

correct

[mild cipher]

and one more question

[nimble crypt]

sure

[mild cipher]

i have trouble in math

and i wanna know ur secret

in learning knowing math so well

and how to become a pro

God tier Pro

without forgetting

[nimble crypt]

im not that much of a pro man i just made a mistake earlier 😅

[mild cipher]

yea, no one is perfect

but u know math very well

[nimble crypt]

remember the methodology, save it somewhere

[mild cipher]

methodology, whats this

[nimble crypt]

how you solve each exercise

[mild cipher]

alr

got it

i'll save it

[nimble crypt]

i.e. remember the log laws same them somewhere and practice until you learn them by heart

[mild cipher]

imma try next question

[nimble crypt]

cause when you forget the methodology you get stuck on the same things and it is tiring

[mild cipher]

alr

[nimble crypt]

i gtg man

gl

[mild cipher]

alr

bye

[nimble crypt]

bye

[mild cipher]

.close

[vale dock]

Channel closed

Closed by @mild cipher

Use .reopen if this was a mistake.

[vale dock]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[mild cipher]

help

[mild cipher]

hello

hello

it says

evaluate $2log_5 x+3log_3 y=8, 6log_5 x - 2log_3 y = 2$

[rocky lotus]

Communist Africa

[mild cipher]

so i tried using elimination method but didn't work out

[midnight pier]

can't understand ur tex writing

[mild cipher]

hello

[normal hollow]

R these two equations?

[mild cipher]

yep

seperately

[normal hollow]

They r asking u to find out x and y?

Alr name the equations

1st one is eqn 1 and 2nd one eqn 2

[mild cipher]

but we were thought to use elimination becaue its short and simple

[normal hollow]

Yes we will do exactly that

Do uk how to?

[mild cipher]

im not gud with it

but i tired then stoppe

[normal hollow]

Alr so the intention of elimination is to eliminate a term with either x or y to find the value of one of them

Alr i'll let u choose which one i want to eliminate

[mild cipher]

x

enemy of all kids since jss1

[normal hollow]

Lol alr

So in order to eliminate x ,we have to ensure that the amount of terms of x is same in both eqn

While u do it ,u have to check if thee coeffecients of the x terms r same in hoth eqn or not

[mild cipher]

nope they're

[normal hollow]

If they r u dont need to do anything and straight up subtract(if both x terms r positive) and add(if they r negative)

[normal hollow]

Yes so we need to do configuration

Now u need to check which of the x terms has higher coeffecient

[mild cipher]

eq 1

[normal hollow]

And wht is it

[mild cipher]

$3log_3 y$

[rocky lotus]

Communist Africa

[normal hollow]

Wait we r talking abt x

[mild cipher]

oh

[normal hollow]

U said u wanted to eliminate x ,no?

[mild cipher]

yea

i wanted

[normal hollow]

Alr so tell me the coeffecients of x terms both eqn

[mild cipher]

ok

$2log_5 x$

[rocky lotus]

Communist Africa

[mild cipher]

$6log_5 x$

[rocky lotus]

Communist Africa

[normal hollow]

Good

Next step is to check whether the coeffecients r divisible or not

[mild cipher]

when were done i'll like to ask u a question

[normal hollow]

Alr

Back to ques

R theey divisible?

[mild cipher]

hm

yea

[normal hollow]

Which one is larger

[mild cipher]

this one

[normal hollow]

Yes so can u multiply something with the smaller one to get the larger one

[mild cipher]

well

[normal hollow]

(When u notice they r divisible,u can definitely do that )
(But if they r not u will have to multiply both equation with each others' coeffecients)

[mild cipher]

im suppose to multiply 6 with eq 1 then 2 with eq 2

[normal hollow]

Nope

That's unnecessary

[mild cipher]

hm

[normal hollow]

For example if ur eqns had 2 and 5 as coeffecient

U k that these rnt divisible

[mild cipher]

yea

[normal hollow]

So u should multiply 2 with 5 amd 5 with 2

[mild cipher]

yea

[normal hollow]

But when it is 2,6 multiply 2 with 3 gives u 6 ,u dont need to do anything with 6

[mild cipher]

wait

brb

im back

sung ji-woo has arrived

[normal hollow]

Hah lol

Alr

[normal hollow]

So do we agrèe to this

[mild cipher]

nope

explain pls

[normal hollow]

Alright

So if u multiply eqn 1 with 3 we get 2×3=6 as coeffecient of x term

Am i wrong?

[mild cipher]

so its not a must we do 2*6

[normal hollow]

No

But it's a step shorter

[mild cipher]

alr

all we have to do is to make it equal

[normal hollow]

Yes

[mild cipher]

just by multipling it once

[normal hollow]

Fiest show me the eqn

[normal hollow]

Remember to multiply the whole eqn

Not js the x term btw

[mild cipher]

here''

[normal hollow]

No i meant after multiplying

[mild cipher]

oh

wait what abt 6

math ehn\

one day i will become better et it

i wont need any bodies help one da

just make math i'll surpass u

[normal hollow]

Sorry phone had shut down on me

[normal hollow]

What

[normal hollow]

Sure keep it up

I mean u hv to multiply eqn 1 with 3 to get something to eliminate

[mild cipher]

alr

[normal hollow]

Multiply the eqn 1 by 3 and show me

[mild cipher]

alr

through out

the eq 1

[normal hollow]

Hm?

Oh yes multiply the whole eqn 1 with 3
Lhs and rhs included

Show me once u r done

[mild cipher]

alr

im done

$$6log_5 x + 9log_3 y = 24$$

[normal hollow]

There should be a plus

And u didnt multiply the rhs with 3

8 needs to be multiplied as well

[mild cipher]

wait

[rocky lotus]

Communist Africa

[normal hollow]

U got it

Now deduct eqn 2 from equation 1

[mild cipher]

alr

[normal hollow]

When u deduct u deduct both lhs amd rhs btw

[mild cipher]

alr but the 2log3 (Y)

would be in negative

[normal hollow]

When u subtract it wouldnt be

[mild cipher]

oh

[normal hollow]

Like 9 log3(y)-(-2log3(y))

[mild cipher]

like from
$$6log_5 x -2log_3 y = 24$$

[rocky lotus]

Communist Africa

[normal hollow]

U can do tht

[mild cipher]

alr

[normal hollow]

But subtracting this from other one would be easier

[mild cipher]

ok

$$11_3 y = 22$$

[rocky lotus]

Communist Africa

[mild cipher]

my answer

[normal hollow]

Yes ood

Shift the 11 to rhs

[mild cipher]

wait were exchanging it

[normal hollow]

There is a log btw

[mild cipher]

so it will -22 = - 11_3 y

[normal hollow]

Logs

Mention them

[normal hollow]

And yes

U r ryt

[mild cipher]

wait

[mild cipher]

y tho

[normal hollow]

Like i said u can do both

[mild cipher]

ok

[normal hollow]

This would be the result of the other alternative

[mild cipher]

i did it without switching

so $$log_3 y = 2$$

[normal hollow]

That's fine

[rocky lotus]

Communist Africa

[normal hollow]

Yes

Log3(y)=2×1

Now express that 1 as a logarithm of base 3

(Man gottta be quick,my time is running out)

[mild cipher]

wait im i not suppose to use log _3 2

$$log_3 y = log_3 2$$

[rocky lotus]

Communist Africa

[normal hollow]

No

U cant write 2 as log3(2)

[mild cipher]

hm

hello

[normal hollow]

So log3(y)=2×1

I termed it wrong

1 should be expressed as logarith of base of 3

Not 2

[mild cipher]

as i said

[normal hollow]

Yes but u xant write 2 as log3(2)

It's wrong

[mild cipher]

how

[normal hollow]

See 1=log3(3)

Do u agree

[normal hollow]

Getting to that(let's be quick)

[mild cipher]

ok

[normal hollow]

So log 3(y)=2log3(3)

Do we agree?

Log3(y)=2×1=2log3(3)
Since 1 =log3(3)

[mild cipher]

yea

[normal hollow]

Now 2log3(3)=log3(3)²

Taking the coefficient to power

Now eliminate log3

What do u get

@mild cipher

[mild cipher]

wait

y = 6

[normal hollow]

3²=6??

[mild cipher]

i did

[normal hollow]

3²=3×3=??

[mild cipher]

y = 2 * 1
y = 2log_3 3
y = 2 * 3
y = 6

[normal hollow]

No u cant do that

Remember we shifted coeffecient to power

[stoic adder]

log₃(3) ≠ 3

[mild cipher]

no wait

there's still log at 2log_3 3

its jsut short form

[stoic adder]

Please use LaTeX if you can

do you mean $2\log_3(3)$?

[rocky lotus]

EightyEightEpsilons

[stoic adder]

Because in that case it would be 2•1, in which that would equal 1

[normal hollow]

Uh when u eliminate log ensure that they dont have coeffecients before log

[mild cipher]

log _3 y = 2log_3 3

[normal hollow]

We made 2×1=2log3(3) cuz that's what we needed

[stoic adder]

Huh?

[normal hollow]

Ig he'll take over

[mild cipher]

like that log_3 cancel the log_3 remaining 3

[stoic adder]

Does 𝑦 = 3² or something?

[stoic adder]

That’s not correct

[mild cipher]

how would i explain

alr whats the answer then

[stoic adder]

$2\log_3(3)=2$, because $\log_{x}(x)=1$, for all $x>1$

[rocky lotus]

EightyEightEpsilons

[stoic adder]

$2\log_3(3) \ =2\cdot1 \ =2$

[rocky lotus]

EightyEightEpsilons

[stoic adder]

understand?

[mild cipher]

yea

[stoic adder]

is there something I’m missing?

[normal hollow]

The main question

Yall r basically heading back to the question

[stoic adder]

I’m doing my best with what I’ve got — I only come here because I stumbled upon something that was wrong

[normal hollow]

Log3(y)=2log3(3)

Make him realize why y=3²

[stoic adder]

Okay

[normal hollow]

Ik ik ,thanks for taking over,igtg

Sorry if it came out as something rude,i didnt mean it that way

[stoic adder]

$\log_a\left(b^c\right)\Longrightarrow c\log_a(b)$

[rocky lotus]

EightyEightEpsilons

[mild cipher]

like i did $$log_3 y = 2;
log_3 y + 21;
log_3 y = 2log_3 3;
cancel log_3;
y=23;
y=6;$$

[stoic adder]

We’re assuming 𝑦 = 3² correct?

[rocky lotus]

Communist Africa

[stoic adder]

Hold up

I’m confused

are we trying to find the value of 𝑦?

[mild cipher]

im showing how y = 6;

[stoic adder]

that second step is wrong

You cannot do that

[mild cipher]

hm

[stoic adder]

$\log_3(y)=2\Longrightarrow y=3^2\Longrightarrow y=9$

Understand?

@mild cipher

[mild cipher]

wait

imm read

[stoic adder]

it’s 9

[mild cipher]

imma read

ok

so the anwser is 3

[stoic adder]

uhh no

[rocky lotus]

EightyEightEpsilons

[mild cipher]

wait

[stoic adder]

WAIT

hold on

DAMNIT

[mild cipher]

see

its 3

[stoic adder]

No it’s 9

[mild cipher]

the square cancels square

[stoic adder]

I wrote the math wrong

[mild cipher]

how

[stoic adder]

your not suppposed to do a square root at all

[rocky lotus]

EightyEightEpsilons

[mild cipher]

thxs guys

[stoic adder]

Use .close once you have figured everything out

[mild cipher]

ok

thxs guys

.close

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Channel closed

Closed by @mild cipher

Use .reopen if this was a mistake.

[mild cipher]

.reopen

[vale dock]

✅

[mild cipher]

what