help-43

[daring axle]

not possible

a/b = sqrt(2)

square both sides

a^2 = 2b^2

the left side is odd but the right side is even

so there's clearly a parity mismatch

[drifting drum]

yeah i should've tested this myself before asking

[daring axle]

lol

[drifting drum]

thanks tho

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[barren verge]

hello guys :)

[barren verge]

how can i integrate this

my prof gave the ans but not the workingsT.T

do i multiply in the factor?

[untold otter]

yes, multiply it in and then it's power rule

[barren verge]

ohh okay tq

damn it looks more complicated than it is

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[kind glen]

how to solve this ?

[vivid breach]

what pattern does the numerator have?

[kind glen]

9-7 = 2, 13-9= 2^2 and so on...

[vivid breach]

do 5S - S to simplify the series

then of the new series, do 5(that) - (that) again

it will be geometric

[kind glen]

i tried S - S/5

[vivid breach]

strange

[kind glen]

ok

[kind glen]

i see

alright solved

thanks

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[deft needle]

take derivative and set it = 0?

dy/dx

im confused how to solve for x with no calculator tho

[thorny heath]

You can write y in terms of x

[spark folio]

the equation looks like a circle that bit a very sour lemon

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[deft needle]

I need help

[deft needle]

No

[pastel cairn]

with?

[deft needle]

I dont get this

[pastel cairn]

where r u stuck at

[deft needle]

dy/dx

[deft needle]

y(y^3+1)^1/2 = x

right

and we do dy/dx

product rule and chain rule

[pastel cairn]

im not so familiar with calc, lemme search up the condition of existance for derivatives

its probably related to the root as were not considering the complex plane tho

[deft needle]

okay

[pastel cairn]

oops its not

the derivative doesnt exist when its = to 0

[deft needle]

yes

[pastel cairn]

find ur dy/dx

and = to 0

[deft needle]

Ok but the issue is

Idk how to find it

[pastel cairn]

then solve for ur y variable

[deft needle]

like i'm getting a whole line of equation for dy/dx

i got dydx(y^3+1)^1/2 + y(1/2(y^3+1)^-1/2)*3y^2(dy/dx)=1

[pastel cairn]

howd u get there

[deft needle]

i dervied it

derived it

[pastel cairn]

u got the steps u used to get there?

it should just be a multiplication and power rule no?

[deft needle]

i ddi the product rule and the chain rule

[pastel cairn]

wait

u did dy/dx

or dx/dy

[deft needle]

yes

dy/dx

[pastel cairn]

that would be abit more complicated i believe

if u just take y as ur variable instead of x itd be simpler

[deft needle]

so d/dx

[pastel cairn]

dx/dy

u just changed variables

instead of f(x) ud have f(y)

[deft needle]

uhh

but im not too sure about f(y)

[pastel cairn]

its the same thing

its just a diff letter

these are how graphs usually are

u take in a function, plug in x and u get ur y

but in our case wed invert it

so we plug in y and get our x

were doing that because taking the dy/dx is more complex in this case than just dx/dy

because X is alone and ready to be worked with, so we just call x our function

[deft needle]

Umm

I think that's too advanced

[pastel cairn]

its not

ur just swapping variables

[deft needle]

Why is it easier than

cuz isn't it the same thing then

just x becomes y

[pastel cairn]

nope

yeah

yeah

but if u were dy/dx

ud have to isolate y

so we just dx/dy instead

because x is already isolated

[deft needle]

okk

Wait

can we use algebra isolate y first

before we derive

That should be easier right

[pastel cairn]

thats what were avoiding

no

if u do that its gonna be more complicated

[deft needle]

ok

dx/dy =

[pastel cairn]

so u take the dx/dy derivative

(same process, just different variables)

[pastel cairn]

there u go

now u just equal it to 0

and find ur Y value

[deft needle]

y=0

no

y=-1

[pastel cairn]

x = 0

and then u solve for that

[deft needle]

y=-1?

omg

it cannot be that

is there a way to do this

[pastel cairn]

yes

[deft needle]

instead of plugging in random numbers

[pastel cairn]

its just algebra now

just a heads up u can group based in sqrt(y^3 + 1)

[deft needle]

is there a quicker way to do this

i'm supposed to do each queestion in like 2 mins

[pastel cairn]

nope

shouldnt be so hard to group that tho

after grouping its just a simple cubic

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[maiden geode]

I know to find sin2theta value for where r is negative

But I find that from -pi to -pi/2 r is positive

And same for pi to 1.5 pi

<@&286206848099549185>

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[inland kite]

um the graph is wrong, its a polar equation, not the normal ones with x and y

ur graph is valid if y = sin2x

its gonna be like a flower

for the polar equation

[maiden geode]

Yah I know that but instead of tabulating theta values I am drawing sin graph to show that r changes when theta changes

[feral flicker]

I think there is a typo in the problem. It should probably say r>0 instead of a>0.

[maiden geode]

No typo my professor told me to take r > 0 for all question

I don’t how that is going to help

[feral flicker]

4a is asking OP to show that the curve C is not in the 2nd and 4th quadrant which is obviously not possible with the given constraints in the problem.

It only makes sense if a>0 is supposed to be r>0.

Here we can see that r is negative.

And negative in the 2nd quadrant here.

[maiden geode]

Thank u very much.

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[harsh prairie]

Hi

[harsh prairie]

I'm m new to the channel and to calculus

[past ether]

Hello whats ur question

[harsh prairie]

I'm having trouble understanding differentistion

[past ether]

!original

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[past ether]

do you have the example problem?

is a lot more easier to target your questions

[past ether]

hmm alr

[harsh prairie]

Question A: verify the point A lies on the curve

[timid anvil]

isn't that just substituting the coordinates of point A into the equation and see if both sides are equal?

[harsh prairie]

Yea

[past ether]

but hes tryna ask how differentiation works

[harsh prairie]

Yea

[past ether]

i dont think that question matters about that tho

[harsh prairie]

Yea that what I came for

[past ether]

differentiation basically is finding the tangent line for that particular point

[harsh prairie]

I thought it was like a formula U substitute to find a gradient at any point

Is that wrong?

[past ether]

uh u substitute that is for integration

is legit the whole opposite thing what differentiation tryna do

[harsh prairie]

Ooooh

[past ether]

differentiation is tryna find the instantaneous rate of change

do you still know

delta y/ delta x

basically what u learn in algebra

the secent line

[harsh prairie]

So the rate of change of the gradient?

[past ether]

uh half is correct

when second line 2 points get very close

but not TOGETHER

[harsh prairie]

Or the rate of change which y changes in respect to x

[past ether]

get infinitely close to each other into a single point

hench differentiation have its definition

as

[harsh prairie]

So it's the rate of change in f'(x) as x gets infinitely close to 0

And and I can't rlly comprehend "the point where x gets close to 0")

The teacher didn't teach us limits
We went straight to differenciation

[past ether]

but in calculus

we bypass that

by using

Limits first

before we learn

differentiation

why we learn limits? To bypass the mathematical errors that we would usually see

like

0/0

inf/inf

those kind of stuff

[past ether]

Ahh i see now

but for normal course

you have to know limits first before you went into differentiation

since differentiation is basically solving lim as x approaching inf

[harsh prairie]

Ooooh

Dam my teacher set so much hm

[past ether]

i have a perfect video if u rlly need holdon

[harsh prairie]

And now I need to learn a whole topic too

[past ether]

https://youtu.be/yfajBIaDf1Q?si=6DVoGu2U7a9SXpAp

there are many other ytbers who featuring limits

like 3b1b

[harsh prairie]

Oooh

Ty

[past ether]

gl on learning limits

you will get it soon

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[drifting drum]

yo

[drifting drum]

for last part

i said that the string would carry weight, which would slow down A, and decrease its acceleration

but the answer was that tension would be different

and idk how that relates to part b, like the question asked

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@drifting drum Has your question been resolved?

[drifting drum]

<@&286206848099549185>

[cloud tusk]

What happens to the tension throughout the string when it has a mass?

[drifting drum]

idk

[cloud tusk]

Well basically the tension throughout the string is no longer uniform

Think of it has a point in the string having to support the weight of the string below it as well

So the top end has to support the weight of the entire string below it as well

Which makes for a bigger tension

[drifting drum]

okay

[cloud tusk]

And a point lower has to support less weight, thus there is less tension there

[drifting drum]

okay

[cloud tusk]

Look at how part b was solved. Can you still do it this way knowing the tension is no longer uniform throughout the string?

[drifting drum]

nope

okay

ty

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[uncut cloud]

would the first method be correct, and is it not (or is) a coincidence that i get 144 from both methods, is my reasoning that we have double counted the ways that 3 abcs can be adjacent correct? thanks

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[vocal matrix]

So I'm getting stuck from the get-go

The only way I know of for finding the characteristic polynomial is finding the matrix and calculating eigenvalue t such that det(A-tI) is 0 for matrix A

But I can't figure out how to represent T as a matrix

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@vocal matrix Has your question been resolved?

[timid lodge]

What is the exercise it is referencing?

[timid lodge]

what is 6.1.12?

[timid lodge]

did you do that?

[vocal matrix]

No, it wasn't assigned

[timid lodge]

well that sure is interesting

[vocal matrix]

I guess I'll take a look

[timid lodge]

just remember the eigenvalue equation: Ax = lamba x. this equation is saying that if you apply the linear transformation A to the vector x (in its domain), it is the same as scaling the input by lambda

so you can apply that same principle to your matrix operator

[vocal matrix]

I'm just not sure how I would check the algebraic multiplicity of any eigenvalue

[vocal matrix]

would you know of any method to find the multiplicity in this case

[timid lodge]

yeah, so the first step would be to find all possible eigenvalues

have you done that?

[vocal matrix]

gimme a sec

[vocal matrix]

I have lambda = 0, -1, 1

[timid lodge]

how did you get 0?

[vocal matrix]

only when the input is a matrix full of zeros

I'm not sure if that's trivial

Okay nevermind 0 cannot be an eigenvector (so 0 is not an eigenvalue here) so just -1 and 1

lambda = -1, 1

[timid lodge]

okay sure, yes -1 and 1

so that tells you that the eigenspace is all matrices that satisfy either

X^T = X or X^T = -X

[vocal matrix]

symmetric or skew symmetric

that's where my understanding breaks down

[dire void]

is symmetric matrices (X^T = X), and for -1 it's skew-symmetric ones (X^T = -X). Together, they span the space of square matrices since any matrix decomposes that way. Need help with the next part?

[vocal matrix]

because I know that a matrix is diagonalizable if the eigenvalues have the same geometric and algebraic multiplicities

[timid lodge]

or if the dimension of the eigenspace is is the dimension of the linear transformation yes

[vocal matrix]

Thanks. But I'm just having trouble going from this to solution

[timid lodge]

so you need to find the dimension of the eigenspace

algebraic multiplicity is hard to find in general

unless the transformation is simple and explicit

[vocal matrix]

so I can bypass the characteristic polynomial

[timid lodge]

yes

do you know how to find the dimension of the subspace of symmetric matrices

[vocal matrix]

aren't there two distinct eigenspaces?

[timid lodge]

well technically yes but they're distinct

so if you find the dimension of both and add them then you'll get the dimension of T

oop didnt see that you said distinct, sorry lol

[timid lodge]

but you can do this

[vocal matrix]

Is this the right way?

Basically 1 occupying every space that's not on the diagonal, and the space mirroring it

[timid lodge]

exactly yes, assuming you're constructing a basis

[vocal matrix]

Great. And it's similar but it's negative on the bottom if it's lambda = -1

[timid lodge]

right. but matrices like

1 0
0 0

are also symmetric

[vocal matrix]

oh right, I'm not sure why I excluded the diagonal

[vocal matrix]

okay if I'm not mistaken the diagonal for eigenvalue = -1 needs to be 0

[timid lodge]

yes. also I think you meant for one of the 1's to be positive for each basis vector for lambda = -1

[timid lodge]

since you said this

[vocal matrix]

man I'm just piling up mistakes today

yes, that is what I meant

[timid lodge]

that's math for you, you know what you're doing but writing out something completely different lol

[dire void]

Hi

[timid lodge]

happens to all of us. but yeah so now just count how many are in each eigenspace

[vocal matrix]

so if I have done my math right, there are n(n+1)/2 in the first one and there are (n-1)(n)/2 in the second one

[timid lodge]

correct

so total dimension is what

[vocal matrix]

so the total would be the summation of that which would be n squared

[timid lodge]

yep, and what is the dimension of T

[vocal matrix]

I get what you're hinting at but I want to back up for a moment
What specifically is dimension of T

[timid lodge]

well it's the dimension of V basically

where T: V -> V

I should've been more specific

[vocal matrix]

I mean V has n columns

[timid lodge]

so the best way to think of dimension is to ask yourself how many basis vectors you need to describe any element of V

[vocal matrix]

n, right?

[timid lodge]

Well V is all n x n matrices. a basis vector would be (for a 2x2 matrix)

1 0
0 0

or

0 1
0 0

etc

[vocal matrix]

oh. so then it should simply be n squared, the same as the total dimension of eigenspace

meaning it is diagonalizable

[timid lodge]

there you go

[vocal matrix]

thank you

[timid lodge]

no problem

[vocal matrix]

this just always works as an alternative to the multiplicity method?

[timid lodge]

it does yes, it's the same thing technically. since geometric multiplicity is always less than or equal to algebraic multiplicity, then if GM(eigenspace) = dim(V), AM(eigenspace) = dim(V) also

but if you are given a problem where you can't easily find AM, they probably want you to use the method we just did

[vocal matrix]

Okay I see, that's cool

Thank you

[timid lodge]

np🫡

[vocal matrix]

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[uncut cloud]

how do you do this? this is probably wrong, but i tried letting the function be x^2 + y^2, so x^2 + y^2 = k(x^2 + y^2), so k=1?

[mild sky]

try to plug in specific values for x,y

[uncut cloud]

okok

[mild sky]

are you sure f(x,y) is any function?

because actually in that case I think that k can be anything

[patent yoke]

Writing down a single example is not helpful when you need to find all possible k. f(x,y) = x - y implies k = -1, but that doesn't establish whether other k aside from +-1 are possible too. You should start with just the relation and work from there instead of writing down specific examples.

[uncut cloud]

i see thanks

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[drifting drum]

for the 2nd part

why cant you set the RHS to 3(1-cos2x)

then divide both sides by 1-cos2x

leaving (secx)^2 - 5 = 3

[kind crane]

it's tan^2(x) not tan(x)

[drifting drum]

ah

misread ty

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[formal sky]

I feel like I did something wrong, how do I fix it?

[main pendant]

Hi! You actually haven't done anything wrong :)

The question looks poorly formatted (it doesn't exactly say what t is, i.e whether P is the population at the end of the th year).

You can say that t = 1.6, or - during the 2nd year will the population reach 60.

[formal sky]

I glanced at the answer key once and from my memory I think it said 5.3

but I'm not sure if my memory is wrong, do you have an explanation on why it might be 5.3?

[main pendant]

5.3 is definitely wrong

If you substitute it in:
P(5.3) = 25(5.3) + 20

,calc 25(5.3) + 20

[boreal girder]

Result:

152.5

[main pendant]

You get something that is definitely not 60

[formal sky]

okay thank you!

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[orchid warren]

If tanTHETA = -1/2 for THETA in Quadrant 2, find sinTHETA + cosTEHTA

[orchid warren]

Precalculus Honors

[runic creek]

What have you tried?

[orchid warren]

I don’t know if O or A is negative so I figured I should do Pythagorean theorem but again, I don’t know which value is negative.

[tired bear]

you could visualize it in the unit circle..

[orchid warren]

How can I know if 1 or 2 is negative.

[runic creek]

And another thing you should know, tan=sin/cos

[orchid warren]

Yes, but how can I know which value is negative.

[runic creek]

Do you know what is "the second quadrant?"

[orchid warren]

When x is negative and y is positive on the unit circle.

[runic creek]

Yep, and what is x and y on the unit circle?

x is cos, y is sin

[orchid warren]

Y is sin

[runic creek]

Yep sorry brain farted

Well, now you should know which is negative and which is positive

[orchid warren]

Oh wait you’re right

Hold on lemme finish the problem

Sin = -1
Cos = 2?

Nvm

Sin = 1
Cos = -2?

Yes?

[runic creek]

Remember, sin and cos is always between 1 and -1

So not quite sadly

[orchid warren]

Sin = .5
Cos = -1?

Could also be:

Sin = .25
Cos = -.5

[runic creek]

Sadly there is no value of theta such that this is true

[orchid warren]

So undefined?

[runic creek]

Wdym?

[orchid warren]

Is this problem solvable?

[runic creek]

Yes it is

But you can't just guess and check like this

Use sin^2 + cos^2 = 1

And we know cos = -2sin

[orchid warren]

Where’d you get this from?

[runic creek]

Sin /cos = -1/2

[orchid warren]

Sin^2THETA or just sinTHETA?

[runic creek]

In which formula?

[orchid warren]

This

[runic creek]

It's derived from the Pythagorean theorem, so it's sin^2

[orchid warren]

How do you hold on lemme comprehend this

So cos is neg and sin is pos, right?

Nvm it is

[runic creek]

Yep

[orchid warren]

And how do you know this?

?

[main yarrow]

triangle??

[orchid warren]

Yes? But you don’t put cos and sin on a triangle

[orchid warren]

And it could be this? @main yarrow

[main yarrow]

tanx=perpendicular/base

[orchid warren]

What’s that?

[main yarrow]

like the soh cah toa

opp/adjacent ig

we can find out our hypotenuse

[orchid warren]

What’s a and o then?

[main yarrow]

Take them as 1 and 2

[orchid warren]

You sure?

A is 1?

Nvm I should just continue with other problems.

.close

[compact pewter]

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[compact pewter]

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[peak turret]

Are logic gates studied in maths too?

[peak turret]

.....

[cunning stump]

Yes, it’s called Boolean algebra

[peak turret]

Cool i have a question

What is this square thing

[cunning stump]

Nor maybe

Idk though, I don’t really know the symbols

[boreal dawn]

I think it's duplicating the signal, but that would be odd

[rich pollen]

Which square thing?

[short ferry]

Are you supposed to simplify the circuit or sth like that?

What's the context

[peak turret]

Truth table

[short ferry]

then this interpretation is slightly less odd

[peak turret]

I don't really know what that gate does

[molten coral]

yea, and we can even find examples of this

must be a teaching device

[peak turret]

Whats that?

[short ferry]

that square is just duplicating the signal (most likely)

its same as the other intersection

[peak turret]

Bruh

Nvm

Thanks

[compact pewter]

@peak turret Has your question been resolved?

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[trail hound]

!status

[compact pewter]

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

[young tinsel]

1

i tried casework but i think that's tedious so yeah

[trail hound]

Count the complement

[molten coral]

the case work shouldnt be tedious?

or maybe it is

[kind crane]

Inclusion exclusion should also work

[molten coral]

the case work isnt so bad

[gusty totem]

ok so you can first find all integers divisible by 5, 2, and 3:

1. include all even integers (so 2, 4, 6, ..., 998, 1000; there are 500 integers like this)

then all numbers divisible by 3 are 3, 6, 9, .., 999 but we alr included the even integers divisible by 3, so include alternating numbers divisible by 3:
3, 9, 15, ..., 999

you can treat this like an ap (a = 3, d = 6; 3 + 6(n-1) = 999; 6n-6 = 996; 6n = 1002; n = 167)

so there are 167 integers like this

then for 5, we should not count even or those divisble by 3, so we have 5(5), 5(7), 5(11), and 5 into other prime numbers

not entirely sure so pls check

[kind crane]

Maybe don't give the entire solution away

[gusty totem]

oh

sorry

[somber mirage]

Too complicated. Just note that 2, 3 and 5 are primes, so a number x is divisible by all of them iff it has them as factors. So $x = 235*t$

[boreal girder]

Dedekind

[gusty totem]

but wouldn't that only be numbers divisible by all?

[somber mirage]

yes, by all of them

[gusty totem]

this says or

it shouldn't be divisible by any

[somber mirage]

then the rest of numbers will not be divisible by 2 or by 3 or by 5

[gusty totem]

so you could have 3 * 3 * 5 which you would end up counting

while you should

[somber mirage]

no

[young tinsel]

nah this is wrong it said inclusive

[gusty totem]

oh ok

[molten coral]

this is not right 👀

[gusty totem]

ill edit it

[somber mirage]

how so?

[gusty totem]

done

[molten coral]

your formulation only counts multiples of 30

[somber mirage]

yes. After that the rest of the numbers EXCEPT the multiples of 30 - are the numbers of the question

[molten coral]

but its not right

15 is divisible by 2 or 3 or 5 but its not a multiple of 30

[kind crane]

Yea this is the way to go

[somber mirage]

then the question require a clarification. Are we looking for the numbers that are not (divisible by 2 or divisible by 3 or divisible by 5)?
Or we looking for numbers that are not divisible by 2 or not divisible by 3 or not divisible by 5?

I was considering the later case

[molten coral]

you shouldnt IMHO because it makes the problem trivial

[somber mirage]

okey

sorry for the confusion

[young tinsel]

So the answer is 266, right?

[kind crane]

!show

[compact pewter]

Show your work, and if possible, explain where you are stuck.

@young tinsel Has your question been resolved?

[compact pewter]

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[trail kelp]

Hello, could someone check if this proof looks good please?

\begin{Definition}[Natural numbers]
The set of the \emph{natural numbers} is the set $\bN = \{1,2,3,4,5,\dots\}$.
\end{Definition}

\begin{Theorem}
It is the case that $$1 \cdot 3 + 2 \cdot 4 + 3 \cdot 5 + \dots + n \cdot (n + 2) = \frac{n(n+1)(2n+7)}{6}.$$
\end{Theorem}

\begin{proof}
We proceed by induction.\\

\underline{Base Case.} The base case is when $n = 1$, and 
$1 \cdot (1 + 2) = \frac{1(1+1)(2 + 7)}{6} = 3$ as desired.\\

\underline{Inductive Hypothesis.} Let $k \in \bN$, and assume 
$$1 \cdot 3 + 2 \cdot 4 + 3 \cdot 5 + \dots + k \cdot (k + 2) = \frac{k(k+1)(2k+7)}{6}.$$

\underline{Induction step.} We aim to prove that the result holds for $k + 1$.
That is, we wish to show that 
$$1 \cdot 3 + 2 \cdot 4 + 3 \cdot 5 + \dots + k(k + 2) + (k+1)(k+3) = \frac{(k+1)(k+2)(2k+9)}{6}.$$
To do this, we begin with the expression on the left, we apply the inductive hypothesis to the sum
of the first $k$ numbers, and from there simplify:
\begin{align*}
1 \cdot 3 + 2 \cdot 4 + 3 \cdot 5 + \dots + k(k + 2) + (k+1)(k+3) &= \frac{k(k+1)(2k+7)}{6} + (k+1)(k+3)\\
&= \frac{k(k+1)(2k+7) + 6(k+1)(k+3)}{6} \\
&= \frac{(k+1)(k(2k+7) + 6(k+3))}{6} \\
&= \frac{(k+1)(2k^2+ 7k + 6k + 18)}{6} \\
&= \frac{(k+1)(k+2)(2k+9)}{6} \\
\end{align*}

Therefore, by induction, $$1 \cdot 3 + 2 \cdot 4 + 3 \cdot 5 + \dots + n \cdot (n + 2) = \frac{n(n+1)(2n+7)}{6}.$$
\end{proof}

Particularly, check if the factorization is correct please

[boreal girder]

Mor Bras

[rotund sphinx]

good

[compact pewter]

@trail kelp Has your question been resolved?

[copper sierra]

its good but you shouldnt say "let n in N" before starting induction

n is the variable of induction so it shouldnt sit freely outside the induction framework

(well you used k instead but point still stands)

[trail kelp]

https://tenor.com/view/writing-notes-checklist-spongebob-squarepants-gif-16129592

Thank you for your comments!

.close

[compact pewter]

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[compact pewter]

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[low pulsar]

how do i turn this exponential function into a logarithmic function

[compact pewter]

Please don't occupy multiple help channels.

[low pulsar]

.close

[compact pewter]

Channel closed

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Use .reopen if this was a mistake.

[low pulsar]

j'

.reopen

[compact pewter]

✅ Original question: #help-43 message

[low pulsar]

how do i turn this exponential function into a logarithmic function

y = 2 x 3^x

[azure vault]

What do you mean by "turn into"

Find inverse function?

!xy

[compact pewter]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[low pulsar]

determine the logarithmic function of the exponential functon

y = 2 x 3^x

[azure vault]

So the inverse logarithmic function

[low pulsar]

like for example if u have y = 3^x

[azure vault]

you need to find x = f(y)

[low pulsar]

the logarithmic of it is x = log(3) y

[azure vault]

You mean x = log_3(y)

[azure vault]

But yes

[low pulsar]

like this

[kind crane]

Closed your duplicate

[low pulsar]

i was trying to but its not letting me

[azure vault]

Well start by isolating the x stuff

[low pulsar]

ok

y = 2 x 3^x

y/2 = 3^x

[cloud warren]

use ln to isolate x

[low pulsar]

idk how to use ln but i know log

[azure vault]

So, which log allows you to retrieve x

[low pulsar]

log y/2 = log 3^x

[cloud warren]

it's the same for ln your base is e. thats all

[azure vault]

If you want to use regular log, sure

[low pulsar]

for what number is my base e

[azure vault]

But maybe with the prior example you gave, you can find a more appropriate log to use than just the regular one

[low pulsar]

i only know log

wdym by a more appropriate log

[azure vault]

Well, all logs come with a base

[low pulsar]

yes

[azure vault]

log(x) is just another way to write log_10(x) if I'm correct about your conventions

[low pulsar]

x is the base

[azure vault]

x is is the output

[low pulsar]

oh alr

[azure vault]

$\log_b(x)$ means "log of x base b"

[boreal girder]

Rafilouyear2026

[low pulsar]

yeah

[azure vault]

And so $\log(x) = \log_{10}(x)$

[boreal girder]

Rafilouyear2026

[low pulsar]

yup

[azure vault]

The usual log is base 10

[low pulsar]

but here it must be 3

because its 3^x

[azure vault]

Well, that's up to you to specify

[low pulsar]

y/2 = 3^x

wait

[azure vault]

You could have used any base for the log to apply on both sides.
But, as I suggested, there is a base that is more relevant than the others here

[low pulsar]

when we do log does the x exponent go the other side right

so now its x^(y/2) = 3

does the y/2 also switch sides

[azure vault]

Idk what sorcery you're trying to invoke

[half bloom]

no

[azure vault]

Just apply log_3 on both sides

[low pulsar]

alright

[half bloom]

you want to ifnd hte inverse of this function

wait

[low pulsar]

log 3 on both sides

is 3 the base

log3 (y/2) = log3 (3^x)

[azure vault]

Yes, log base 3

$\log_3(\frac y2) = \log_3(3^x)$

[boreal girder]

Rafilouyear2026

[low pulsar]

yeah

now what

im confused for thius part

[azure vault]

Well

Remember log_3 is the inverse of 3^...

So doing one after the other cancels both actions

[low pulsar]

the x goes the other side

[azure vault]

You end up back at where you started

[low pulsar]

oh

[azure vault]

What other side 💀

[low pulsar]

why did we do that then

[azure vault]

To retrieve x

[low pulsar]

other side of the equality

[azure vault]

$\log_3(3^x) = x$

[boreal girder]

Rafilouyear2026

[low pulsar]

thats what i mean

wait

1 question tho

there was 2 log_3

[azure vault]

?

[low pulsar]

did the x stay there or did it go on the other side of the equality

[azure vault]

This has nothing to do with y. It's just a factual statement

If you take 3^something

And then take the log_3

You get what you started with

$\log_3(3^{whatever}) = whatever$

[boreal girder]

Rafilouyear2026

[low pulsar]

oh i see

x = log_3(3^x)

[azure vault]

So this becomes

[low pulsar]

x = log_3(y/2)

alright

tysm

what if its y = 2 x 3^(x+2) -4

[azure vault]

Same thing as before, try to isolate x

[low pulsar]

y + 4 = 2 x 3^(x+2)

y/2 + 2 = 3^(x+2)

[cloud warren]

it should be y+4

[low pulsar]

oh yeah

log y/2 + 2 = log 3^(x+2)

x + 2 = log_3 (y/2)

[azure vault]

Some stuff disappeared on the y side

[low pulsar]

u mean the y/2?

[azure vault]

No

The +2

[azure vault]

log y/2 + 2 = log 3^(x+2)

[buoyant heath]

Exponential and logs are inverses

Switch the x to y and solve for y

[low pulsar]

x = log_3 (y/2) - 2

[azure vault]

Bruv, as I said, you made a +2 magically disappear

[cloud warren]

u forgot to put the +2

[buoyant heath]

It’s subtracting

[low pulsar]

x + 2 = log_3 (y/2)

the + 2 is on the y side

[buoyant heath]

x - 2 = log_3 (y/2)

oh wait

[azure vault]

There were two +2s

[low pulsar]

oh

[azure vault]

Look at the boldened +2

[cloud warren]

x= log_3 (y/2 +2) - 2

[low pulsar]

ohh

yeah

got it alright thanks

.close

[compact pewter]

Channel closed

Closed by @low pulsar

Use .reopen if this was a mistake.

[low pulsar]

.reopen

[compact pewter]

✅ Original question: #help-43 message

[low pulsar]

determine the rule of the logarithmic function

f(x) = log_c((b)(x-h))

to find b

-3 = 1/b -4

b = 1

?

[kind crane]

Where did 1/b - 4 come from

[bleak flower]

<@&268886789983436800>

[low pulsar]

1/b is in the equation and -4 is the asymptote

1/b - h

[kind crane]

Where is 1/b here

[low pulsar]

to find the value of b the equation is

Zero = 1/b + h

for a logarithmic function

[kind crane]

I see

[low pulsar]

yeah

-3 = 1/b - 4

if no mistakes

[kind crane]

Yea b=1 looks right

[low pulsar]

alright

y = log_c((1)(x-h))

1 = log_c((1)(-1+4)

c = 3

?

[kind crane]

Actually I still think this is a mistake

[low pulsar]

how

[kind crane]

x=-3 and y=0 from the graph. When you plug that in you get
0 = log_c(b(-3-h))

b(-3-h)= 1

is h=-4 or h=4

[low pulsar]

well

im double checking it should be correct

if u got a coordinate like (-3,0)

or (1,5.0)

and u got h = 1

1,5 = 1/b + 1

h = x

x = asymptote

[latent flame]

Yes

[kind crane]

Alright this and the rest look right. I got confused on the sign of h

h=-4 is right so x-h = x+4

[low pulsar]

is b = 1 tho

[kind crane]

Yes

[kind crane]

-3=1-4 checks

[low pulsar]

alright

we have c h and b

f(x) = log_3 1(x+4)

?

[kind crane]

Why do you still have c

[low pulsar]

forgot

should be good

[kind crane]

Yea

[low pulsar]

ty

.close

[compact pewter]

Channel closed

Closed by @low pulsar

Use .reopen if this was a mistake.

[low pulsar]

.reopen

[compact pewter]

✅ Original question: #help-43 message

[low pulsar]

if i have -1 = log_c((1)(2+2))

-1 = log_c (4)

c^-1 = 4?

[low pulsar]

what do i do here

.close

[compact pewter]

Channel closed

Closed by @low pulsar

Use .reopen if this was a mistake.

[low pulsar]

.reopen

[compact pewter]

✅ Original question: #help-43 message

[low pulsar]

i dont understand my mistake i have 2 coordinates

[cold vale]

Yo

[low pulsar]

hey

[cold vale]

I accidentally slept last night my bad

[low pulsar]

(9,0) and (9,9.-1)

[low pulsar]

dw lmao

ty for helping

[low pulsar]

right

and the asymptote is 10

h= 10

first thing i do is find b with the coordinate (9,0)

the equation for it is

[cold vale]

What topic is this

[low pulsar]

Zero = 1/b + h

[low pulsar]

logarithmic

[cold vale]

What's your question

[low pulsar]

i think i did a mistake so just making sure

well i did do a mistake

9 = 1/b + 10

-1 = 1/b

-b = 1

b = -1

?

[cold vale]

I mean that's correct b=-1

Can someone check <@&286206848099549185>

[low pulsar]

its good cause the graphic shows it

b < 0

and in this case b = -1

the next step i did was try to find c

which is where my mistake is

-1 = log_c ((1)(9,9-10))

-1 = log_c(-0,1)

c^-1 = 0,1

[cold vale]

What's your original question

[low pulsar]

its to determine the equation of the logarithmic function

so find b,h,c

we found b and we got h

h = asymptote

[cold vale]

And you use this?

[low pulsar]

yeah to ifnd b

[cold vale]

Can you take a pic of the problem

[low pulsar]

alr

[cold vale]

And what are you finding again

[low pulsar]

c and b

f(x) = log_c((b)(x-h))

-1 = log_c((1)(9,9-10))

-1 = log_c -0,1

c^-1 = -0,1

1/c = -0,1

-10 = c

which doesnt make sense

[cold vale]

C should be 10

[low pulsar]

how

[cold vale]

We have $-1=log_c((b)(9-10))$ right

[boreal girder]

sice19

[low pulsar]

yeah

[cold vale]

Did you already get your b right?

Or not yet?

[low pulsar]

b is 1

[cold vale]

Nope

[low pulsar]

how

oh ur right

[cold vale]

It's -1

[low pulsar]

how

[cold vale]

Don't forget the signs

What's your answer

[low pulsar]

9 = 1/b + 10

-1 = 1/b

-b = 1

@cold vale

[cold vale]

we have f(x)=log_c(b(x-h)) right

[low pulsar]

y7eah

[cold vale]

Use the point (9,0) and x=10

Substitute it

[low pulsar]

i did

9 = 1/b + 10

[cold vale]

We get 0=log_c(b(9-10))

[low pulsar]

-1 = 1/b

[cold vale]

Remember log both sides

So whats log(1)

=0 right

[low pulsar]

yeah

[cold vale]

Then reverse it

So equal to 1

[low pulsar]

log 1 = 0

how do u reverse

[cold vale]

Yes

0=log 1

Then remove log both sudes

[low pulsar]

c^1 = b(9-10)

[cold vale]

Nope

Wait

[low pulsar]

arent we supposed to find b first

[solemn matrix]

$f(x) = \log_c(b(x-h))$

[boreal girder]

1 divided by 0 equals Infinity

[solemn matrix]

i just got in this channel and it's already a chaos

[low pulsar]

to find b right

9 = 1/b + 10

[solemn matrix]

im going to just start over with this
-# (YOU HAVE 3 CHANNELS 😭)

we got $(9, 0)$ on the graph

[boreal girder]

1 divided by 0 equals Infinity

[solemn matrix]

so $f(9) = 0$

[boreal girder]

1 divided by 0 equals Infinity

[low pulsar]

oh yeah about those

i closed them but theyre still there

[low pulsar]

yeah

[solemn matrix]

and what's $f(9)$?

[boreal girder]

1 divided by 0 equals Infinity

[low pulsar]

f(9) is value of y when x = 0

[solemn matrix]

hold on, what's his original question? i don't get where this h comes from

[solemn matrix]

substituta in here

[low pulsar]

h is asymptote

[low pulsar]

k

0 = log_c((b)(9-h))

[solemn matrix]

that's one equation

and then $(9.9, -1)$ is on the graph, so $f(9.9) = -1$

[boreal girder]

1 divided by 0 equals Infinity

[solemn matrix]

and that's 2 equations

[low pulsar]

yup

[solemn matrix]

from that you can solve for $b$ and $c$

[boreal girder]

1 divided by 0 equals Infinity

[solemn matrix]

-# (through h)

[low pulsar]

alright ima try that

[solemn matrix]

also i think the asymtote is at $x = 10$?

[boreal girder]

1 divided by 0 equals Infinity

[low pulsar]

yes

it is

[solemn matrix]

yeah so substitute $h = 10$ too

[boreal girder]

1 divided by 0 equals Infinity

[low pulsar]

for 0 = log_c((b)(9-10))

ncm

when its c = b(9-10)

what do i do here

c = -1b?

[solemn matrix]

hold on

[low pulsar]

kk

[solemn matrix]

so $\log_c(-b) = 0$

[boreal girder]

1 divided by 0 equals Infinity

[low pulsar]

ohh true

b is negative right

[solemn matrix]

and think about this

we know that $c^{\log_c(-b)} = -b$

[boreal girder]

1 divided by 0 equals Infinity

[low pulsar]

yeah

[solemn matrix]

remember that $\log$ and exponents cancel each other

[boreal girder]

1 divided by 0 equals Infinity

[solemn matrix]

and since $\log_c(-b) = 0$, then $c^0 = -b$

[boreal girder]

1 divided by 0 equals Infinity

[solemn matrix]

and the only way that this can happen is when $-b = 1$

[boreal girder]

1 divided by 0 equals Infinity

[low pulsar]

i got this too

but how can it be -b/

its supposed to be b so i switched it

b = -1

[solemn matrix]

okay

so substitute $b = -1$ in the next equation

[boreal girder]

1 divided by 0 equals Infinity

[low pulsar]

kk

wait

c^-1 = -0,1

c = 10

alr tysm

.close

[compact pewter]

Channel closed

Closed by @low pulsar

Use .reopen if this was a mistake.

[low pulsar]

.reopen

[compact pewter]

✅ Original question: #help-43 message

[low pulsar]

to find b in -1 = log_2 ((b)(0-h))

2^-1 = ((b)(0-h))

do i do 2^-1 = 0,5?

or 1/2

[hexed hornet]

as in $-1 = log(2, b (0 - h))$

[boreal girder]

Mirror

[hexed hornet]

?

oops

hold please

[low pulsar]

alright

nvm its the same thing

0,5 and 1/2

[hexed hornet]

the quantity being log_2'd is $(b(0-h))$?

[boreal girder]

Mirror
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[hexed hornet]

lemme go find the original because i am not understanding the original query

$-1 = \log_2(b(x-h))$ is the original query, and we're looking for b?

[boreal girder]

Mirror

[low pulsar]

yeah

[hexed hornet]

ok... what do we know about h?

[low pulsar]

nothing

we have to isolate b

[hexed hornet]

is it a constant or a variable?

[low pulsar]

variable

[hexed hornet]

👍

[dim bridge]

I need help regarding maths project

[hexed hornet]

!occupied

[compact pewter]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[solemn matrix]

!occupied

[compact pewter]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[hexed hornet]

ok so raising to power of 2 is the right call

$\frac{1}{2} = b(x-h)$

seems to me like all we need to do to isolate b is divide by (x-h)

[low pulsar]

wait no

cause it -1 = log_2 b(x-h)

[solemn matrix]

-# h is the asymptote btw

[low pulsar]

so 2^-1 = b(x-h)

right?

[hexed hornet]

yes you're right

[solemn matrix]

hold on a second

[boreal girder]

Mirror

[low pulsar]

yeah

[hexed hornet]

is $x-h$ ever zero?

[solemn matrix]

that's better

[boreal girder]

Mirror

[hexed hornet]

because if it is, we can't divide by it

[solemn matrix]

-# i think we can still divide, but with a twist here

[low pulsar]

we aren't

[hexed hornet]

-# agreed but i'm trying to get them to find that out

[low pulsar]

we making 2 equations

for b

thats the first one

[hexed hornet]

there's another equation?

[low pulsar]

b = 0,5/(0-h)

[hexed hornet]

!original

[compact pewter]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[low pulsar]

yeah i was gonna show it

[hexed hornet]

kk

so the other equation says $b = \frac{-1}{2h}$

[low pulsar]

-2 = log_c (b)(-1-h))

[boreal girder]

Mirror

[hexed hornet]

from this

[low pulsar]

2^-2 = b(-1-h)

[hexed hornet]

ok back up a second you've got too many equations here

[low pulsar]

its the same one im just solving it