How the fuck do I do this

How the fuck do I do this

a*log(x) = log(x^a)
log(x)+log(y) = log(x*y)
log(x)-log(y)=log(x/y)
That's all you need to know

Thanks, user

Don't mention it. Just try and really understand the steps you need to get to the solution.

Final result for comparison is A = log ( (x^2 + (x-1)^3) / (x^2+1)^2 )

Log is not part of the answer

Been trying to, haha, but this semester isn't going to well for me math wise.
If I don't nail the exam tomorrow I'm just gonna drop out of it and try something next year, but I have to do this online work first that half of it doesn't even pertain to the test but it's also due tomorrow

What is your major?

Comp science for the first semester, but since I hate it/coding in general when I thought I would like it, I'm floating more towards something that I'll at least like.
My college is cheap as fuck so I'm not exactly needing a 6 figure salary coming right out of it

Well, it means that you only need to put the log's argument in the box. However, the mathematically correct answer comprises the log!

Log((((x^2+1)/x)^2)(x-1)^3)
Therefore A=[(x^2+1)/x]^2*(x-1)^3

You could simplify this further, but I won't, can't do this shit on a tablet

Oh I get what you mean. Yeah, A is just the argument

Isn't that (x^2+1)/x supposed to be a x/(x^2+1)?

Ignore this, I fucked it up... should be [x/(x+1)]^2*(x-1)^3

Yes.

what a nice boy, one internet for you my friend. (unrelated poster)

I give up.
Forgot the x^2+1 in the denominator.

Gave up right when you had it right!

I'd do anything to avoid writing my thesis

Come suck my dick then faggot.

This is the Sup Forums I know and love. Thanks for the chuckle

Use wolframalpha.com
Or Symbolab.com

Um. Also maybe

Plug into wolfram and see what you get in alternate forms

To start of with:
Ln(x^10(x-1)^1/2)-??

Ok. Same as before, only the inverse process. Take the factors inside the argument of the logarithm (the x^10, the sqrt(x-1) and the (3x-4)) and create three logarithms with those factors as arguments. Remember that when you have a factor at the denominator you need to put a - in front of the logarithm. After you have done this, take the exponents of the arguments of the logarithms and put them in front of the log, using the first one of the properties shown in .

You can do this, just take one step at a time. If you need further explanations just ask

(I meant that you need to use the first property of from right to left: you take the exponent and put it in front of the ln)

Ln(x^10(x-1)^1/2)-ln(3x-4)

That's right, now you can also separate ln(x^10(x-1)^1/2) into a sum of two lns!

Yes.
So now you just do the rules to this one Ln( x^10(x-1)^1/2)

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A=10
B=1/2
C=-1

Boom, headshot. Nice work.

If I go here, I better recieve dick pics

Not the guy posting that link, but I will provide as many dick pics as wanted for helping for my work

Pussy lips for dick pics only.

Samefagging?

I'll take as a payment one facebook or instagram pic of the hottest girl you know for each problem we helped you solve. Possibly in a bikini. Or naked.

Kek this thread. Sup Forums is first class education.

A= sunken battleship