Help me prove a calculus law

[Slater]

For homework, my professor wanted us to prove two things.
1) That a^(x - y) = (a^x)/(a^y)
2) That ln(x^r) = r*ln(x)

The first one was a piece of cake, less than half a page with a superflux of space to spare.

The second one... is meh.

So far, I've proven that (d/dx)(ln(x^r)) = (d/dx)(r*ln(x))
But from there I'm not sure where to go.

Any help?

[Doubleshadow]

What does r* signify? And is this for a proofs class or a calculus class, so I know how to format the answer?

[Slater]

r times ln(x)

We are prooving a calculus law of natural logs

[Sweet Mercury]

For homework, my professor wanted us to prove two things.
1) That a^(x - y) = (a^x)/(a^y)
2) That ln(x^r) = r*ln(x)

The first one was a piece of cake, less than half a page with a superflux of space to spare.

The second one... is meh.

So far, I've proven that (d/dx)(ln(x^r)) = (d/dx)(r*ln(x))
But from there I'm not sure where to go.

Any help?

Do you have to prove it using calculus methods? Because you should be able to prove that logorithmic property using only precalculus mathematics.

[Doubleshadow]

Alright, if you are just showing one side is equal to the other, I have that. If you need a step by step break down of a proof in written English that'll be a different story.

(d/dx)(ln(x^r)) = (d/dx)(r*ln(x))
Take the derivative of both sides and using the Power Rule (don't forget r is a constant and does not change):
(1/ (x^r))[r(x^(r-1))] = (r/x)
Rewriting it, no algebra here:
[r(x^(r-1))]/(x^r) = (r/x)
Subtracting the bottom of the first half of the equation from the top of the first half of the equation to simplify:
[r(x^(-1))] = (r/x)
Rewritting again:
(r/x) = (r/x)

[Slater]

yep, that's as far as I got, but that doesn't exactly prove it. Derivitives that are equal can differ by a constant.

[SP1]

Take the exponent of both sides:

e^(ln(x^r)) = e^(r*ln(x))

resolving the left hand side:

x^r = e^(r*ln(x))

Now in the left side, Z^(a*b) = (Z^a)^b...

x^r = (e^(ln(x))^r = x^r

Hope this helps...

SP1

[Dante]

yep, that's as far as I got, but that doesn't exactly prove it. Derivitives that are equal can differ by a constant.

Actually it does, note that the constant must be zero, you definied it to be at the beggining of your proof.

1) That a^(x - y) = (a^x)/(a^y)
2) That ln(x^r) = r*ln(x)

note that if you integrated double shadow's proof to get back to the original equations, than you would have an added constant C, but they must be zero to equal your original question.

Hope that makes things easier,

Pascal

[Icarus]

If I may argue with Pascal, what Doubleshadow did is not what he seems to think. She proved that dy/dx g = dy/dx h. That does not imply the equality of g and h. For example,

f=3x, g= 3x+1.

f' = g', but f != g.

SP1 has a nice answer, and here is mine:

ln (x^r) = r( ln x)
ln (x*x*x*x*...*x) = r( ln x) with x repeated r times.
(ln x) +(ln x)+ ... +(ln x) = r(ln x) rules for logarithms.
r(ln x) = r(ln x)

[Dante]

hmm AHH its late... Gomenosai!! eto... sorry but I gues I did mess that up... Either way I use this same sorta logic in physics all the time anyways... constants additions don't matter for the most part because you can just define most of them away to zero.

sorry,
Pascal

[Slater]

what I did was integrate the derivative of ln(x^r), which gives you rln(x) + C. Since R is a constant as is C, you can solve for C for plugging an arbitrary value in for x. Thus, I chose 1... so...

r*ln(x) + C -> C = -r*ln(x) -> C = -r*ln(1) = 0

Thus, since the constant was 0, it can be concluded that ln(x^r) = r*ln(x)

Now, on to calculating volumes of weird shapes...

[Doubleshadow]

Dang it! Stupid C. Guess who was always losing unnecessary points on her exams in calc. 2 that way. I was always forgetting that... I'm glad you got it. I had my text book from my proofs class, but not my calculus classes or I would have not forgotten that part.

[Icarus]

Dang it! Stupid C. Guess who was always losing unnecessary points on her exams in calc. 2 that way. I was always forgetting that...

I did that too. It got so bad that I started to just go down the page and put +C at the end of all the problems at the start of the test.