help-17

[sage monolith]

B is supposed to be the "background knowledge". Or known information that is already taken for granted.

[spiral turtle]

I mean, based on the assumptions given, yes the math follows, provided you assume n < 10^136, which as I pointed out is at least a little bit suspect.

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[twin meteor]

Halex

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[steep lintel]

Can someone help verify if these partial derivatives and if my integrals are right

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[fluid comet]

idk i try to simplified it but i couldnt

[silk osprey]

what’d you get

when simplifying

[fluid comet]

8?

[silk osprey]

well i meant your simplified expression

without the logs

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[peak tundra]

shaded region

[idle robin]

big circle area - small circle area - big circle 40 degrees area + small circle 40 degrees area

[idle robin]

or
big circle area - 40 degrees big circle area - 320 degrees small circle area

do whatever u understand

[peak tundra]

how do that tho

big circle raidus is 5m

[idle robin]

u have radius use area formula

[opaque shell]

which method is more intuitive to you

and we can explain from there

[idle robin]

theta /360 * pi*r^2 area when angle is given

[peak tundra]

big circle = 78.54

small = 28.27

[opaque shell]

okay okay

that looks good

[peak tundra]

40/360*78.54?

[opaque shell]

okay so what happens if we subtract them

78.54 - 28.27

what area would we get

[peak tundra]

50.27

[opaque shell]

mhm

but what does this area represent

[peak tundra]

the ring

[opaque shell]

mhm

so we have the ring

[peak tundra]

but we have minus the cutout

[opaque shell]

exactly

hoew would we do that

[peak tundra]

how fidn area of that

[opaque shell]

well we have that the angle inside is 40 deg

[peak tundra]

yes

[opaque shell]

if we were to go all the way around

how many degrees would that be

[peak tundra]

360

40/360?

[opaque shell]

so we did 40/360 degrees

mhm

[peak tundra]

0.11

[opaque shell]

mhm

so what can we do with that

[peak tundra]

0.11* 78.54

?

[opaque shell]

perf

but that's over

we need to subtract the area in the smaller circle too

remember

[peak tundra]

yes

is it 38.44m^2

50.27-(8.717+3.109)

[opaque shell]

I believe so

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[modern storm]

1, 1, 1, 2, 2, 3, 4, 5, 7, 9, 12, 16, 21, 28, 37, 49, 65, 86, 114, 151, 200, 265, 351, 465, 616, 816, 1081, 1432, 1897, 2513, 3329 ( padovan's spiral sequence )
1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096 ( 2^n )

how do i prove that the highest amount of numbers i will need from padovan's spiral sequence to reach one of the numbers from 2^n is 4?

1 (2^0) can be "reached" with 1 (first number from padovan's spiral sequence)
2 (2^1) can be reached with 1 + 1 (second and third)
2 > 2+2
4 > 3+4
8 > 7+9
16 > 12+16
32 > 21+28
64 > 37+49
128>65+86
256 > 114+151
512 > 200+265+351
1024 > 465+616
2048 > 816+1081+1432
4096 > 1897+2513

[modern storm]

also know that padovan's spiral number sequence is defined as: $p\left(m\right) = p\left(m-2\right)+p\left(m-3\right)\$ with $p\left(1\right) = 1$, $p\left(2\right) = 1$ and $p\left(3\right) = 1$

[twin meteor]

zzz0nnn

[modern storm]

i tried this in desmos with logarithmic type, it seems both have exponential growth, but 2^n seems to be faster, reaching about double of Padovan's Spiral

im not to keen on how logarithmic graphs work, but would that mean that the distance becomes bigger and bigger, which means that theres no max amount but rather just gets bigger over time?

oh, i gave desmos three times the dots for padovan's function

and it surpasses 2^n, so it should be lower than b, maybe.

ill multiply for bigger amounts

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[wary fiber]

found w in terms of e

now wht to do

w= root 3/2+ i/2= cos pi/6+ n sin pi/6= e^ipi/6

[wary fiber]

Multiple choice btw

[lone linden]

Uhhh whoever wrote this didn’t do a good job b/c the answer isn’t unique

[vocal sleet]

@wary fiber Has your question been resolved?

[wary fiber]

?

Multiple options are correct

Jee advanced question, they are never incorrect

( in most cases)

[lone linden]

Ah. As a word of advice, say “multiple correct” - most ppl here interpret “multiple choice” as there being multiple options to pick from (aka not what you meant)

Anyway, think about what the real part of z^n is. You should get that there’s two possibilities to consider - ||when the real part is -sqrt(3)/2 or -1 for H2|| and ||sqrt(3)/2 or 1 for H1||. You can then find the angle in these ||4|| cases.

[wary fiber]

Sorry my bad

I'm having trouble finding the angle

How do you find the angle

[lone linden]

Draw it out and consider ||adding the angles from z1 and z2||

Also, word of advice (again) - say what you’ve gotten up to so helpers don’t go over certain parts unnecessarily (it does get annoying after a while)

!status

[vocal sleet]

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

[lone linden]

Use this as a guide if you have to

[wary fiber]

Alright I will choose I don't know where to beign 1.

If there is a mistake in my calculation

[lone linden]

Oh I meant that you would say something like “I got up to finding the values in H1 and H2, but idk how to find the angle”

1 is you starting from square zero

[lone linden]

Well pick one of the cases

What does the diagram look like

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[mystic talon]

There are 4 sections on a math paper. Section A and section B has 2 questions respectively, section C and section D has 4 questions respectively, altogether 12 questions on this paper.
Now how many ways to do 6 questions if every candidate has to choose at least 1 question but 2 questions at most from each section?

[wary fiber]

Nvm I got it

[mystic talon]

I have three ideas on this question

1st idea
1st step: I do 1 question from each section first, which is 2×4×2×4=64 ways
2nd step: I do another 2 questions from the rest of the questions
So there are 3 cases:
1st case is I pick the questions from section A and B, which will be 1×1
2nd case is I pick 1 question from section A/B, and then pick another 1 question from section C/D, which will be 2×1×2×3=12
3nd case is I pick the questions on section C and D, which will be 3×3=9
So there are 1+12+9=22 ways
And altogether I have 64×22=1408 ways

2nd idea
1st step: I do 2 questions from each section, which is 2C2×2C2×4C2×4C2=36 ways
2nd step: I remove 2 questions I did from 2 sections, which will be 4C2×2×2=24
And altogether I have 36×24=864 ways

[lone linden]

This overcounts by a bunch because of the double counting in the sections you pick two questions from

[lone linden]

I don’t understand what the second step is supposed to do here

[mystic talon]

Why would this step make me overcount tho

[mystic talon]

Um my aim is to do 8 questions and then just cancel 2 extra questions to only answer 6 questions

[lone linden]

See edit

[lone linden]

I was referring to the expression itself

Where did those values come from

Because it should be split by cases or subtracting another complement imo

[mystic talon]

Um so do I just need to split into 6 cases

[lone linden]

Depends on how you count cases ig

You can group some of them together

1. You pick two questions from A and B and one question from C and D
2. You pick one question from A and B and two questions from C and D
3. You pick two questions from (A or B) and (C or D) and one question from each of the other sections

This is how I would do it

But doing 6 cases based on the (4 choose 2) possibilities for which sections to pick two from isn’t wrong

It’s just more tedious

[mystic talon]

Ooo and that will get me a value of 352?

[mystic talon]

Um how do I overcount tho
Assuming I pick 2 questions from A and B, and 1 question from C and D

[lone linden]

In this method, the order in which you pick the questions in the sections you do two questions in matters

It shouldn’t

[mystic talon]

Mmm

What would the correct calculation be for that step

Oh wait I think I get how it overcounts

[lone linden]

It’s more like the premise as a whole is off

It can be fixed but it’s not the ideal way

,w 64((1/2)^2 + (3/2)^2 + (1/2)(3/2)*4)

[twin meteor]

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Upgrade to WolframAlpha Pro!

Requested by civil_service_pigeon

[mystic talon]

Oh my god this looks hard to figure out💀

[mystic talon]

1st step: I do 2 questions from each section, which is 2C2×2C2×4C2×4C2=36 ways
2nd step: I choose 2 sections, and then remove 1 question from the sections I pick, which will be 4C2×2×2=24
And altogether I have 36×24=864 ways

[lone linden]

That doesn’t answer my question

[mystic talon]

Um

[mystic talon]

Oh that
I choose 2 sections from the 4 sections, which is 4C2
And then from the 1st section I choose, I remove 1 question from the 2 questions I did, so there are 2 ways
And from the 2nd section I choose, I remove another 1 question from the 2 questions I did, so 2 ways too
And then it would be 4C2×2×2=24

[mystic talon]

My idea of this step is like
There are 4 groups and every group has 2 apples
And then I take 2 apples from the 8 apples but the 2 apples cannot be from the same group

[lone linden]

This overcounts again

Ex. Suppose you pick questions 1 and 2 in section C. Then, removing question 2 will leave you with just question 1. However, the result is the same if you start with 1 and 3 and scrap 3

[mystic talon]

Oh

So this is also not the ideal way to approach this question

[lone linden]

Nah

[mystic talon]

One last question, may I ask how to get this calculation

[lone linden]

You’re just accounting for the overcount factor in each case

[mystic talon]

Ooo do we learn that on uni

[lone linden]

Uh idk

It’s the easier end of high school combi for me

[lone linden]

1. 1/2 comes from the fact that the method says that there’s 2 ways to pick two questions from A/B when it should be 1. You’re working with A and B, so square it.

2. The given method multiplies by 3 to get another question from C/D, but this overcounts by 2, so multiply by 3/2. Square for the same reason states above.

3. 1/2 and 3/2 explained earlier, 4 comes from there being two ways to pick one of A/B and two says flick one of C/D

[mystic talon]

Oh I think I get it

Thank you so much! You helped me a lot

.close

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[late mirage]

ive got no idea how to do this

its in a circles + hyperbola chapter and i havent learnt it...

[opaque shell]

oh

you can distribute

x^2(x+y^2) - y^2(x+y^2)

[bronze osprey]

what's the question?

[late mirage]

oh graph it

mb

[opaque shell]

I think simplify LHS

ohhh

[bronze osprey]

ahhhh so zero product property, x^2 - y^2 = 0 OR x + y^2 = 0

for first one use difference of 2 squares

the second one is y^2 = -x

that's the reflection of x^2 = -y or y = -x^2 across the line y = x

(we swapped x and y)

[late mirage]

so it should be a line and a parabola?

[bronze osprey]

x^2 - y^2 = 0 is two lines

and x + y^2 = 0 is a parabola

[late mirage]

OH yes

ok i think ik how to do it now

dont know why its in this chapter tho...

ook i got it

thank u

.close

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[merry imp]

hi

[merry imp]

how did

the variables in the top become like inthe bottom

[opaque shell]

we subtract on both sides

wym?

[merry imp]

am talking about -ypower2 -2x-y

we brought them to the other side so they became negative ik that

but am confused on the order

shouldnt it be -ypower2-y-2x

instead of -ypower2-2x-y

[opaque shell]

does the order matter?

[merry imp]

it does or it doesnt?

[opaque shell]

y^2 - y - 2x = y^2 - 2x - y

no it doesn't

[merry imp]

ok then finish it off with me

[opaque shell]

-y^2 - y - 2x = -(y^2 + y + 2x) = -(y^2 + 2x + y) = -y^2 - 2x - y

maybe that can help you see why more clearly

[merry imp]

in the first statemenet we take y as a coomn factor so it become y'(x+x2y)=-ypower2-y-2x

i need to find the value of y'

my points are (1,1)

[opaque shell]

perfect

divide by (x+x2y)

[merry imp]

like this right?

[opaque shell]

looks about right

[merry imp]

listen to make it easier for you

am just confused on wtf my teacher did here

at the start of the second page when she begins substituting

the denominator suddenly the positive became a negative

is she wrong or is it supposed to be like that?

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[merry imp]

.close

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[torpid steppe]

can someone please help me solve (2x+1)/4-x/3=4

[torpid steppe]

I keep getting the wrong answer

[river minnow]

Show your work

[torpid steppe]

I changed it to a common denominator /12

so it was (6x+3)/12 -4x/12 = 4

which got me 2x+3=4

which meant that 4-3=2x so 2x=1 and x=0.5

[river minnow]

Where did /12 go?

[torpid steppe]

OH

so it would be 2x+3/12 =4?

[river minnow]

(2x+3)/12 = 4, yeah

[torpid steppe]

2x=22.5

sorry

I eman x=22.5

mean

yep thats righ

t

thanks for the help! have a great day/night

.close

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[haughty rover]

time is x axis and distance is y axis im assuming

wait

is it the other way around

distance is x-axis and time is y-axis

[grizzled vault]

independent variable in x axis

[haughty rover]

ya so distance

wait

yeah distance

right?

[grizzled vault]

time is independent

distance is influenced by time here

[haughty rover]

right

[haughty rover]

but like this this context, we're trying to analyse speed in a distance-time graph

are u 100% sure

[grizzled vault]

so what?

[haughty rover]

fair point

😭

ok so time x-axis and distance y-axis

[grizzled vault]

if u don’t believe me just google “distance-time graph”

and check the axes

[haughty rover]

yes sir

now

how should i develop my functions

[grizzled vault]

do you know interpolation?

[haughty rover]

nope

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[minor walrus]

quick q, could anyone explain why its B but not D? im not sure why

[modern ledge]

looks like hsc ex2?

doesnt rly matter who

*tho

what do u notice about the velocity vector

to be more specific, do you recognise that the velocity vector with x dot and y dot represents the direction of air resistance in this case

if air resistance is proportional to speed squared

u need a direction

[minor walrus]

yeah it is

[modern ledge]

which is why we have magnitude speed = v

multiplied by direction

[minor walrus]

2022

[modern ledge]

d will give you velocity cubed

u have to remember both x dot and y dot are speed but with a direction

[minor walrus]

ohhh my god

i see now

i kinda forgot that v = x dot, y dot

😭

[modern ledge]

v = |x dot| yes

trick question that probably would have tricked a lot of ppl

best of luck for ur task tmr

[minor walrus]

yeah

thank you 🙏

[modern ledge]

honestly just get in some good sleep

im heading bed early tonight for tmrs exam

[minor walrus]

yeah im prob gonna sleep at 12

not gonna bother staying up all night

[modern ledge]

dont we start at 1:30

[minor walrus]

got early entry so im not too stressed anyways

[minor walrus]

think so

1:50?

iirc

[modern ledge]

oh right

my school just wants us there earlier ig

makes sense

anyhow, won't occupy u any longer

best of luck

[minor walrus]

alr

ty

gl for tmr

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[wind orchid]

which is the correct answer for "25,200 at 2% per semi annual for 6 months, find the simple interest"

252 or 504?

[mellow raven]

Simple interest formula = P x r x t
P = 25,200

r = 0.02. (2% is 0.02)

t = 0.5 years (6 months = 0.5 years)

[wind orchid]

I'm confused for n

it says semi annually but time is only 6 months?

so t should be 1?

since it only occured once a year

disregarding the semi annually in the given

then I should get 252 if I'm not wrong?

[mellow raven]

t = 1

yes

its semi anual

so twice a years

[wind orchid]

okay so 252 is the correct answer

thank you

.close

[mellow raven]

wait am confused

i find 504

[wind orchid]

it is 504

😂

but, isn't this a dumb question right

Cuz why do they put semi annually when t is 6 months

It's impossible

[chrome juniper]

ahh, another australian studying for the test tmr haha

oop, sorry. THought that was the current problme mb

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[fading cypress]

is this correct?

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[plush summit]

quick question, when I have a function f(x) = -x² and I insert 4

is it then -(4)² = -16 oder -4² = 16?

[robust spruce]

-(-4)²=-(16)=-16

[plush summit]

-16 oh i checked it

okaaay

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[vocal sleet]

@vivid eagle Has your question been resolved?

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@vivid eagle Has your question been resolved?

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@vivid eagle Has your question been resolved?

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[vast shale]

help me please

[vast shale]

how is angle a

angle b

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[sleek edge]

Hi, this is a physics problem but i am not confused about the physics but rather the math operations used here

[sleek edge]

im confused about the second step going from (c^2-u^2)(c^2+Ux'V)^2 in general tbh

How did the Ux get cancelled out when there are no other Ux present in the equation

this solution was done in class but im trying to work it out myself rn

[vast shale]

Hmm this looks a bit complicated

Perhaps they expanded the whole thing and then came to this simplification?

Cause intuitively, I can’t see how this can be formed

[sleek edge]

it should be the expansion of (-u^2)(c^2+ux'v)^2 right?

like this

since c^2 is already accounted for there

I gave up on figuring it out

This is part two to that question

Would this make sense as an answer? Or is there a better way to answer this

[vocal sleet]

@sleek edge Has your question been resolved?

[sleek edge]

someone help pls 😭

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[slender lintel]

Hello how do you solve this equation in complex numbers. I am seriosly stuck only managing to rewrite the right-hand side to polar and reducing it

[viral copper]

that's about it

Let $2 + i = w$, then:
[ e^{2z} = 2 + i \implies 2z = \log (2 + i) = \ln \abs{w} + i(\arg (w) + 2k \pi)]
where $k \in \set Z$

[twin meteor]

krypton

[slender lintel]

is this a formula and if so what's it called?

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[slender lintel]

Thank you so much for the help ^^

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[viral copper]

.nah you just write it in polar form and use properties of the logarithm

[jagged cargo]

\

[jagged cargo]

sorry

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[sudden vine]

Suppose that a 5x5 matrix is known to have exactly two distinct eigenvalues, 4 and 9. If the trace of the matrix is 40 and the determinant is 26,244 then what is the multiplicity of the eigenvalue 4?

[sudden vine]

Hey guys I need help in this question, can someone show me how to do it please?

[hard atlas]

the trace is the sum of the eigenvalues and the determinant is the product of the eigenvalues

[sudden vine]

oh so then 40 is sum of all the eigenvalues of that 5x5 matrix?

is it like for example x is the multiplication of eigenvalue 4 and y is for 9, then 4x + 9y = 40?

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[bitter pilot]

What have you tried

[vast shale]

j (x) = log5 (x + 5)
y = log5 (x + 5)
x = log5 (y + 5)
....

[bitter pilot]

Ok good

Now watch the base

It's 5

So the trick is to do on both sides 5^(...)

[vast shale]

(5) x = log5 (y + 5) (5)

[twin meteor]

bacc (unhelpful)

[vast shale]

j (x) = log5 (x + 5)
y = log5 (x + 5)
x = log5 (y + 5)
(5) x = log5 (y + 5) (5)
5^x = 5^log5 (y + 5)

[bitter pilot]

Yea

Now watch this

[twin meteor]

bacc (unhelpful)

[bitter pilot]

The bases cancel

[vast shale]

j (x) = log5 (x + 5)
y = log5 (x + 5)
x = log5 (y + 5)
(5) x = log5 (y + 5) (5)
x = 5^log5 (y)?

[twin meteor]

bacc (unhelpful)

[bitter pilot]

So what remains is

[twin meteor]

bacc (unhelpful)

[vast shale]

j (x) = log5 (x + 5)
y = log5 (x + 5)
x = log5 (y + 5)
(5) x = log5 (y + 5) (5)
5x = logy?

[bitter pilot]

Whats inside log remains

[vast shale]

j (x) = log5 (x + 5)
y = log5 (x + 5)
x = log5 (y + 5)
(5) x = log5 (y + 5) (5)
5^x = 5^log5 (y + 5)
5^x = y + 5?

AHHH WAITTT

j (x) = log5 (x + 5)
y = log5 (x + 5)
x = log5 (y + 5)
(5) x = log5 (y + 5) (5)
5^x = 5^log5 (y + 5)
5^x = y + 5
5^x - 5 = y??

[loud walrus]

🥳

Not 5x btw

5^x

[bitter pilot]

Yea you did it!

[vast shale]

1/7^2x = 49
7^-1(2x) = 49
7^-1(2x) = 7^2
-1(2x) = 2
-2x = 2
x = 2 + 2?

<@&286206848099549185>

[pine hull]

-2x = 2
Try dividing both side by -2

[vast shale]

1/7^2x = 49
7^-1(2x) = 49
7^-1(2x) = 7^2
-1(2x) = 2
(-2) -2x = 2 (-2)
x = 2 (-2)?

[pine hull]

Dividing not multiplying

-2x/-2 = 2/-2

[vast shale]

1/7^2x = 49
7^-1(2x) = 49
7^-1(2x) = 7^2
-1(2x) = 2
-2x/-2 = 2 /-2
x = -1?

[pine hull]

Yup

[vocal sleet]

@vast shale Has your question been resolved?

[vast shale]

100 = 2^x+5 - 28
2² × 5² = 2^x+5 - 2² × 7?

[pine hull]

Try adding 28 both side first , it will be easier to work with then

[vast shale]

100 = 2^x+5 - 28
100 - 28 = 2^x+5
2^3 × 3^2 = 2^x+5?

[pine hull]

100+28 = 2^x+5

[vast shale]

100 = 2^x+5 - 28
100 + 28 = 2^x+5
2^7 = 2^x+5

I can do it on my own now

[vocal sleet]

@vast shale Has your question been resolved?

[vast shale]

Idk what to do

[wind geyser]

,w 1/8

[twin meteor]

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[torn timber]

step 1:write 0.0625 as a fraction

[vast shale]

okeh

[wind geyser]

,w 1/16

[torn timber]

its ||1/16||

[wind geyser]

well F

[vast shale]

(0.0625)^x = 2^x+12
1/16^x = 2^x+12
Okay I think I can do this on my own now

[wind geyser]

yes

[torn timber]

good :)

[vast shale]

Okay no I don't

(0.0625)^x = 2^x+12
1/16^x = 2^x+12
16^-1(x) = 2^x+12
2^4(-1)(x) = 2^x+12
4(-1)(x) = x+12
...

[pine hull]

4(-1)(x) is -4x
-4x = x+12
I think you can solve now

[vast shale]

(0.0625)^x = 2^x+12
1/16^x = 2^x+12
16^-1(x) = 2^x+12
2^4(-1)(x) = 2^x+12
4(-1)(x) = x+12
-4x = x+12
-4x-x = 12?

[pine hull]

Yeah -4x-x = -5x

[vast shale]

What- how did that happen

[pine hull]

Because it's subtraction, for an example
-4-1 becomes -5

So -4x-x = -5x

[vast shale]

Ohhh

(0.0625)^x = 2^x+12
1/16^x = 2^x+12
16^-1(x) = 2^x+12
2^4(-1)(x) = 2^x+12
4(-1)(x) = x+12
-4x = x+12
-4x-x = 12
-5x = 12
X = -12/5?

[pine hull]

Yeah

[wind geyser]

do u know how logs work?

[vast shale]

no...

[wind geyser]

well

[torn timber]

apply log_e on both sides kek

[wind geyser]

¯_(ツ)_/¯

[torn timber]

logs are basically the inverse of exponentials

[vast shale]

log_e(5)=x

[wind geyser]

damn

yes

[torn timber]

yes

[wind geyser]

how do u not know logs then 😭

[vast shale]

Wait let me identify...

[vast shale]

I didnt listen to the...

[wind geyser]

!

noice

[fickle dock]

its just log 5

[vast shale]

5 = e^x

e is the base
5 is the number inside the parenthesis
x is... just left alone since it's just x and y???

so..

log_e (5) = x?

[fickle dock]

thats all

[wind geyser]

that's correct

[fickle dock]

ln 5 is all

[wind geyser]

it says your answer must be in log_b(a) = c

here b = e, a = 5 and c = x

[vast shale]

So.. in an exponent...

(a =) 5 = (b =) e^(c =) x

Correct?

[fickle dock]

What the fuck is this notation?

lol

[wind geyser]

uhhh

[vast shale]

Sorry

[wind geyser]

wdym lmao

you mean to say a is 5?

b is e

and c is x

right?

[torn timber]

[wind geyser]

such that $\log_b(a)=c$?

[twin meteor]

@wind geyser

[vast shale]

5 = e^x

a = 5 =
b = e^
c = x

[wind geyser]

ah

well ^ is not a different thing

it just means you're raising to a power

superscript

[torn timber]

the power is not required here

[fickle dock]

Let me try and explain it. Logarithms and roots. and Raising to the power. are all definable in terms of each other. They're like three connected operations.

https://www.youtube.com/watch?v=sULa9Lc4pck

[wind geyser]

dam

[torn timber]

roots are in a way raising to a power log are inverses of raising to the power in a certain sense dont worry abt a lot kek

[vast shale]

So if I answer this...

21 = 4^x + 5
21 - 5 =4^x
16 = 4^x
log_4 (16) = x?

[wind geyser]

yes

[torn timber]

yes

[vast shale]

oh

WHY DID I HAVE A HARD TIME IN THE QUIZ

THIS IS EASY

[torn timber]

it happens to the best of us

[fickle dock]

Cool you play chess.

[vast shale]

Embarrassing note I was only 6th place in chess

AAAAAAAAAAAAAAA

I HATE IT SO MUCH

[fickle dock]

6th place out of how many.

[vast shale]

There are 31 in our compettition

[fickle dock]

you Would destroy me

lol

[torn timber]

if you are done close the channel

[wind geyser]

you're ahead of 25 people in the competition

[vast shale]

NOT YET

[wind geyser]

isnt that great

[vast shale]

PLEASE IM SORRY

[vast shale]

IM NOT EVEN 3RD PLACE

[fickle dock]

Okay, continue continue. We can continue talking about chess and dms if you want later.

[wind geyser]

next time

[vast shale]

IT'S HORRIBLE TRULY HORRIBLE

[wind geyser]

there's always the next time

[torn timber]

there is #chess-go-shogi

[vast shale]

So this is like a reversed question

So log_1(5) = 0?

[fickle dock]

If you are Asian this confirms a stereotype.

[wind geyser]

log_1 does not work

[vast shale]

Mabuhay

[vast shale]

It should be above 1?

[fickle dock]

yes

[vast shale]

Hmm..
Log_5(0) = 1

I think since it should be above 1 it's suppose to be 5? 0 is the closest and 1 is.. just there

[torn timber]

log_1 does not exist

[vast shale]

How about this?

[fickle dock]

No, you can't do that. We just told you it should be above one.

I'm actually getting actively confused by your notation. Could you just send the problems you have?

[vast shale]

Just confirmmmmm

log_5(0) = 1?

[fickle dock]

I'm not even sure what that means. Where's the =

[vast shale]

WAIT ABOVE 1 BASE IS SUPPOSE TO BE ABOVE 1 AND THE THINGY IN THE PARENTESIS CAN BE 1

log_5(1) = 0???

[torn timber]

log_5(1)=0 lol

[vast shale]

I got it >:3

[torn timber]

good :3

[fickle dock]

just think of it this way. Logarithms help you find the exponent. You're looking for the Exponent.

[torn timber]

i suggest you see a vid on logarithms it will be helpful

[vast shale]

I got it

[fickle dock]

You can look at the one I sent you in the thing. If it doesn't help you, you can try other ones.

[vast shale]

Now... base is 7
(x-4) put on the other side
5 can be exponent?

[wind geyser]

Try applying 7^... to both sides

and notice that $7^{\log_7(\ldots)}=\ldots$

[twin meteor]

@wind geyser

[vast shale]

log_7 (x-4) = 5 (7)
7^log_7(x-4) = 7^5?

[torn timber]

Note do this only in the reals

[wind geyser]

thats right, now 7^log_7(anything) is = anything

[torn timber]

Yes now simplify the LHS

[vast shale]

LHS??..

[fickle dock]

Left hand side.

[torn timber]

Left hand side

[vast shale]

(x - 4) = 7^5?

[torn timber]

Yes

[wind geyser]

yes

[vast shale]

That's the answer?

[wind geyser]

well no

it asks u to write the answer in $b^c=a$

[twin meteor]

@wind geyser

[torn timber]

It will be better if you add 4 on both sides

[vast shale]

7^5 = (x - 4)?

[wind geyser]

nop

add 4 on both sides

although

,w 7^5+4

[twin meteor]

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Upgrade to WolframAlpha Pro!

Requested by shr1han

[wind geyser]

how would one write this lmao

[vast shale]

but.. it's looking for the equation..?

Are you sure?

[wind geyser]

yeah..

maybe you're right

put that

[vast shale]

Okeh

Is log = 10?

[fickle dock]

Do you have to find eggs? Or do you have to rewrite it?

X

[wind geyser]

well

b is 10

[wind geyser]

when you don't have something in the bottom as b

its by default 10

(in this context)

[vast shale]

Where did you get 10-

[vast shale]

oh

[wind geyser]

do the same thing now: 10^...

both sides

[fickle dock]

Are you sure? Maybe log is by default e Probably you're right.

[wind geyser]

hmm..

[wind geyser]

have you heard of something like ln? or a constant called e? euler's constant?

[vast shale]

What the..?

[fickle dock]

I think ln might be e and then log might be 10.

[vast shale]

Okay this is making me anxious if you say "might"

[wind geyser]

$\ln(...)=\log_e(...)$

[twin meteor]

@wind geyser

[vast shale]

I don't get it

[wind geyser]

ln is just a log with b = e

another notation for log_e is ln

[vast shale]

So.. what he said In is e then log is 10

[wind geyser]

ln = log_e

log = log_10

[vast shale]

Ohhhh okay

[torn timber]

Usually this kind of distinction is made in non math textbooks

[wind geyser]

o

[torn timber]

Usually if you read math books they write log while the actually mean ln

[wind geyser]

here im taught that in chemistry log() is the base-10 one

[wind geyser]

in physics and math log is ln

[wind geyser]

do you know factoring?

[vast shale]

Yes-

[wind geyser]

$a^2-b^2=(a+b)(a-b)$

[twin meteor]

@wind geyser

[wind geyser]

use this to factor the left hand side

of the equation

[fickle dock]

Wait, the question is already solved.

[vast shale]

log_5(625x)^2-64 = 0
hmm...
625x^2 + (5 - 64)^2 = 0
625x^2 - 59^2 =0
(25 + 1) (25 - 59)?...

[fickle dock]

Already have an answer plugged in into that picture you sent.

[torn timber]

i would go with another approach

[vast shale]

So.. you just want me to answer
a^2 - b^2 = (a + b) (a-b)

Im sorry I don't.. really think teacher would identify a number that way??

[torn timber]

i would rather do this
(log_5(625))^2=64
applying square root on both sides
log_5(625x)=\pm 8

[vast shale]

Wouldn't that be confusng?

[fickle dock]

No, he's right. His is the more simple way.

[torn timber]

it is the same thing

[vast shale]

log_5(625x)^2-64 = 0
log_5(625)^2=64
log_5(25x)=8

[fickle dock]

No, no, no, no.

[vast shale]

then that means you didn't square root both sides you only square root the right side?

[fickle dock]

Your square rooting both sides because you remove the exponent

5^2=25

Now watch what happens when I square root both sides.

5=5

[vast shale]

Bruh make it specific

[fickle dock]

I did make it specific. I gave you a specific example.

[wind geyser]

welp yeah

[vast shale]

Well mak it math language

[fickle dock]

It is math language, okay, I'm sorry I can't help. Somebody else needs to try to explain it.

[wind geyser]

i was making it more specific about the +- part

[vast shale]

It may be understandable for you but look how I understand it

[torn timber]

it is the same lol

[vast shale]

I didn't know that was gonna be how it's gone

[wind geyser]

lets start from the first

(log_5(625x))^2=64 do you get this much?

[vast shale]

log_5(625x)^2-64 = 0
log_5√(625)^2=√64 <- Isn't this easy in the eyes than just typing it? Because all I was thinking was "Oh square root everything" of course my interpretation will be like √log_5(625)^2=√64

[wind geyser]

._.

[fickle dock]

√log_5(625)^2=√64

[wind geyser]

do you understand that log_5(625x) is a whole quantity thats squared?

[fickle dock]

This is the correct interpretation.

[vast shale]

You can't square root log or 5

[wind geyser]

why not

[log_5(625x)]^2 = 8^2

square root

you get

log_5(625x) = +- 8

[vast shale]

There's no square root for 5 than 5 square root 1 and log will just be log.

[wind geyser]

thats what you do, right?

[vast shale]

I follow it very specifically

[wind geyser]

see this, you have a big quantity named log_5(625x) which is squared

This big quantity could have been anything

[vast shale]

Meaning if you just said "square root" to a person that don't know much about logs then I have to follow it very specifically to know what happened

[wind geyser]

log_5(625x) is just any number

[torn timber]

$(\log_5{(625x)})^2=64\
\sqrt{(\log_5{(625x)})^2}=\sqrt{64}\
\log_5{(625x)}=\pm 8$

[twin meteor]

convergence

[wind geyser]

👍

this

[fickle dock]

x^2=9 Here's the simplest example. the way you would find X here, if you square root both sides. and you'll see on the left all that happens is the exponent disappears. So you're left with X. And on the right you're left with the square root of. 9 which is +_/three. So you have X = -+3.

[vast shale]

So meaning yes although "√log_5(625)^2=√64" correct in your eyes the process in my head will go √log√5√(625)^2 = √64

[wind geyser]

lol

you don't know about logs and you're solving these questions

sooo

[fickle dock]

What the hell? Why did you add more square roots? That's not how it works. lololol

[wind geyser]

we do assume you know how to write log notation at least

[vast shale]

That's literally what I'm saying I don't know much about the process so if you don't tell me specific instruction it'll be incorrect for you 😭

[fickle dock]

OK, let's should we try another approach.

[vast shale]

Let's.. just move on...

log_5(625x)^2-64 = 0
log_5(625)^2=64
log_5(625x)=8
what do I do now

[fickle dock]

Think about it this way, every time you square root something, all you're doing is dividing the exponent by two.

[vast shale]

okie

[wind geyser]

Do you know log properties?

well i assume no

[vast shale]

well... like I said didn't listened to the teacher...

[wind geyser]

epic

so

first thing:

$\log_b(ac)=\log_b(a)+\log_b(c)$

[twin meteor]

@wind geyser

[wind geyser]

Try to simplify the left side using this

[torn timber]

lol i think it will be better if you know what log is ,it will really help you

[wind geyser]

^^^^

you could also at this point do 5^... on both sides but the numbers will get too big

[vast shale]

log_5(625x)^2-64 = 0
log_5(625)^2=64
log_5(625x)=8
log_5(625x)(8) = log_5(625x) + log_5(8)?

[wind geyser]

no no

log_5(625x)

here your "ac" is 625x, that is, a = 625 and c = x

[vast shale]

Idk which one is a and which one is c

[torn timber]

.

[vast shale]

What part of TELLING ME SPECIFICALLY YOU GUYS DON'T UNDERSTANDDD 😭

[wind geyser]

i did tell specifically

also when you did this this isn't valid

i only said to simplify the LEFT HAND SIDE

[fickle dock]

we Can try a different approach log_(a)(a^b)=b Use this.

[vast shale]

I didn't know which one is a and c

[wind geyser]

sure

now you do

continue

[vast shale]

Of course I'm going to guess that it's a number

[wind geyser]

i did say to see the left hand side only

which means the right must be left intact

well either way

my bad

continue

[torn timber]

ok so $625x =625 \times x$

[twin meteor]

convergence

[vast shale]

Well I didn't know that x should be BLEK moving onnn

log_5(625x)^2-64 = 0
log_5(625)^2=64
log_5(625x)=8
log_5(625)(x) = 8
og_5(625) + log_5(x) = 8

What now

[wind geyser]

do u know what log_5(625) is?

Think of raising 5 to SOME power to get 625

[vast shale]

625 x 5?

[wind geyser]

what is that SOME power

noo

[vast shale]

625^5?

[wind geyser]

5^SOMETHING = 625
find SOMETHING

[vast shale]

OHHH

[wind geyser]

Do you know anything related to exponents about 625

[vast shale]

5^4

[torn timber]

lol you didnt need to shout

[wind geyser]

oh

yes

corerect

[wind geyser]

eeeee

[torn timber]

yes

[wind geyser]

yes

[wind geyser]

So, log_5(625) is 4

[vast shale]

What-

how-

[torn timber]

thats how log function works 💀

[wind geyser]

When you say log_5(625) = something

you mean to say

5^something = 625

both things are exactly the same

since you found something = 4

log_5(625) = something = 4

[vast shale]

Something like...
5^_ = 625

the answer is 4?

[wind geyser]

tbh, dont take it to heart, but you really should learn logarithms from youtube

[wind geyser]

yes

[vast shale]

So Log_4

[wind geyser]

no

the entire log_5(625) is 4

replace log_5(625) with 4

[vast shale]

Oh

log_5(625x)^2-64 = 0
log_5(625)^2=64
log_5(625x)=8
log_5(625)(x) = 8
log_5(625) + log_5(x) = 8
4 + x = 8?

[torn timber]

indeed mathcord is not the best place to learn log

[torn timber]

its 4+log_5(x)=8

[wind geyser]

real

[vast shale]

log_5(625x)^2-64 = 0
log_5(625)^2=64
log_5(625x)=8
log_5(625)(x) = 8
log_5(625) + log_5(x) = 8
4 + log_5(x) = 8
What do I do now

[wind geyser]

subtract 4 from both sides

and do the usual business of solving log_b(a) = c

[vast shale]

log_5(625x)^2-64 = 0
log_5(625)^2=64
log_5(625x)=8
log_5(625)(x) = 8
log_5(625) + log_5(x) = 8
4-4 + log_5(x) = 8-4
log_5(x) = 4

[torn timber]

yes

[vast shale]

now what

[torn timber]

5^(log_5(x))=5^4

[fickle dock]

use this

[vast shale]

log_5(625x)^2-64 = 0
log_5(625)^2=64
log_5(625x)=8
log_5(625)(x) = 8
log_5(625) + log_5(x) = 8
4-4 + log_5(x) = 8-4
log_5(x) = 4
5^log_5(x) = 5^4
625

[fickle dock]

yes

[vast shale]

YIPPEEE

[wind geyser]

yeah that's all

[torn timber]

thats one solution x=625

[wind geyser]

true.

[vast shale]

Idk what to do with this

[wind geyser]

Divide both sides by 2

[fickle dock]

OK, this one is like one of the easiest ones.

[wind geyser]

and do it yourself next

we will not say

[vast shale]

Okie

2log_2 (x + 1) = 4
(x + 1) = 2?

[wind geyser]

wrong

[vast shale]

2log_2 (x + 1) = 4
log_2 (x + 1) = 2?

[wind geyser]

correct

[torn timber]

yes

[fickle dock]

that is the correct first step Yes.

[wind geyser]

now what do u think u should do

you have done this twice

or thrice

[vast shale]

Square root both sides

[wind geyser]

in this session

nop

think again

[torn timber]

no..

[wind geyser]

what did you do when you saw log_b(...) = c

[fickle dock]

this demonstrates that you don't even grasp the basic concept of what a logarithm is. You really should just watch a video on it before you continue.

[wind geyser]

did you... do something with exponents?

[vast shale]

2log_2 (x + 1) = 4
log_2 (x + 1) = 2
log_2 (x) + log_2 (1) = 2?

[wind geyser]

no...

[fickle dock]

Just read the Wikipedia article please.

[vast shale]

2log_2 (x + 1) = 4
log_2 (x + 1) = 2
√log_2 (x + 1) = √2

[fickle dock]

log_a(B)=X a^X=B

Look at the equation I just posted and think about it carefully. This is what a logarithm is.

[vast shale]

2log_2 (x + 1) = 4
log_2 (x + 1) = 2
log_2 (1) = x
2^x = 1

[fickle dock]

no

[torn timber]

no

[fickle dock]

Look at the equation I just posted and rewrite it.

[vast shale]

2log_2 (x + 1) = 4
log_2 (x + 1) = 2
2^2 = (x+1)

[fickle dock]

yes

[vast shale]

2log_2 (x + 1) = 4
log_2 (x + 1) = 2
2^2 = (x+1)
4 = x + 1
4 - 1 = x
3 = x

[fickle dock]

yes

[fickle dock]

I would advise you to write this down somewhere and think about it carefully.

[vast shale]

Divide both sides by 5?

[torn timber]

remember that $a^{\log_a{(...)}}=...$

[twin meteor]

convergence

[vast shale]

5^log5(x+5) + log(1000) = 16

5^log5(x+5) + log(1000) = 16
5^log5(x+5) = log(1000) + 16

[torn timber]

no

[wind geyser]

wrong

What do u think $\log(1000)$ is?

[twin meteor]

@wind geyser

[wind geyser]

I remind you these facts:

[vast shale]

5^log5(x+5) + log(1000) = 16
...

[wind geyser]

1. $\log(\ldots)=\log_{10}(\ldots)$ for this context

[twin meteor]

@wind geyser

[vast shale]

I don't think I could remember it with these heavy shaking and heart beat

5^log5(x+5) + log(1000) = 16
5^log5(x+5) + log10 (10) = 16

[wind geyser]

2. $\log_{b}(b^{\text{something}})=\text{something}$

[twin meteor]

@wind geyser

[wind geyser]

what's wrong?

[vast shale]

It's normal due to pressure and stress

[wind geyser]

alright

[vast shale]

5^log5(x+5) + log(1000) = 16
5^log5(x+5) + log10 (10^3) = 16

[wind geyser]

no..

10^something = 1000

what is something?

[vast shale]

5^log5(x+5) + log(1000) = 16
5^log5(x+5) + log10 (10^3) = 16

[wind geyser]

yes!

now use those properties

we have used this lot of times now

to be specific:

[vast shale]

What do I do now

[wind geyser]

1. $a^{\log_a(\ldots)}=\ldots$

2. $\log_a(a^{\ldots})=\ldots$

[twin meteor]

@wind geyser

[wind geyser]

Try applying these here

[vast shale]

5^log5(x+5) + log(1000) = 16
5^log5(x+5) + log10 (10^3) = 16

1. 5^log5(x+5) = ...
2. log10 (10^3) = ...

[wind geyser]

well

for the first term

your "..." is x+5, right?

[vast shale]

yes

OHHHH

[wind geyser]

so replace 5^log5(x+5) with x+5

similarly, for the 2nd term

replace log10 (10^3) with 3

[vast shale]

5^log5(x+5) + log(1000) = 16
5^log5(x+5) + log10 (10^3) = 16

1. 5^log5(x+5) = (x+5)
2. log10 (10^3) = 3

[wind geyser]

yes the two points are correct

now replace the Left Hand Side terms with those points

[vast shale]

5^log5(x+5) + log(1000) = 16
5^log5(x+5) + log10 (10^3) = 16

1. 5^log5 = (x+5)
2. log10 (10) = 3

[wind geyser]

no...

[vast shale]

hmm...

[wind geyser]

that numbered list i gave you is for your knowledge

[wind geyser]

1. and 2. here are correct, try using this information in your equation

5^log5(x+5) + log10 (10^3) = 16
this equation

[vast shale]

5^log5(x+5) + log(1000) = 16
5^log5(x+5) + log10 (10^3) = 16
5^log5(x+5) = (x+5); log10 (10^3) = 3

[wind geyser]

well

yes

but i want you to replace the terms in the equation

5^log5(x+5) + log10 (10^3) = 16

replace each of the things (separated by +) with the information you know

[vast shale]

5^log5(x+5) + log(1000) = 16
5^log5(x+5) + log10 (10^3) = 16
5^log5(x+5) + log10 (10^3) = 16 (x+5) (10^3)

[wind geyser]

no..

I said

the left hand side

has two things

5^log5(x+5) and log10 (10^3)

But we know that

5^log5(x+5) is just x+5

and log10 (10^3) is just 3

can you simplify the left hand side

[vast shale]

https://tenor.com/view/aqua-aaaaa-cry-help-anime-gif-24539456

[wind geyser]

lmfao

chill

[vast shale]

5^log5(x+5) + log(1000) = 16
5^log5(x+5) + log10 (10^3) = 16
x+5 + 10^3 = 16

[wind geyser]

not 10^3

your ... part was 3

so just 3

[vast shale]

5^log5(x+5) + log(1000) = 16
5^log5(x+5) + log10 (10^3) = 16
x+5 + 3 = 16

[wind geyser]

Yes

now can you solve it?

its now a simple linear equation

[vast shale]

5^log5(x+5) + log(1000) = 16
5^log5(x+5) + log10 (10^3) = 16
x+5 + 3 = 16
x+5 = 16 - 3
x + 5 = 13
x = 13 - 5
x = 8

[wind geyser]

yes!

thats it

[vast shale]

Idk how to do this

[wind geyser]

ah

okay

so

there is another property you need to use

$\log_b(a)=\frac{\log a}{\log b}$

[twin meteor]

@wind geyser

[wind geyser]

Can you use it in the left side of the equation?

You have two logs multipled

log_x(5) by this property can be written as what?

[vast shale]

(log_x 5) (log_5 x) = log_7 x
log_x 5 = log_5/log_x
log_5 x = log_x/log_5
log_7 x = log_x/log_7

[wind geyser]

i said, left side of the equation

keep the right side as it is

alright so we know that:

log_x 5 = log5/logx
log_5 x = logx/log5

can you replace the things in the left side with the known information?

[vast shale]

(log_x 5) (log_5 x) = log_7 x
(log_5/log_x) (log_x/log_5) = log_7 x

[wind geyser]

not the _

they are just numbers inside logs

(log(5)/log(x)) * (log(x)/log(5)) = log_7 x

[vocal sleet]

@vast shale Has your question been resolved?

[silk rampart]

Hello, do you know of the property log_b (a) = 1/log_a (b)

[vocal sleet]

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[plush summit]

hi, I wanted to check if my f'(x) is correct or not

in the picture f(x) is given

[plush summit]

my f'(x)

[river minnow]

,w differentiate (4x^3*ln(x))/(x^3-1)

[twin meteor]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by alonerbean

[plush summit]

I'm not sure if I also need to multiply out (x³-1)²

[river minnow]

Your answer is correct

[river minnow]

You can check if your derivative is correct using wolframalpha btw

[plush summit]

Okay thank you, and is this f'(x) also correct?

[plush summit]

thank you, I will look into it

[river minnow]

That's correct too

[plush summit]

,w differentiate 5cos(-3x + 1/6 pi) * e^-4x^2-1

[twin meteor]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by skenox

[plush summit]

bruh

[river minnow]

Make sure to use parenthesis

,w differentiate 5cos(-3x + 1/6 pi)e^(-4x^2-1)

[twin meteor]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by alonerbean

[plush summit]

Okay thank you:D

.close