log function e^(ln12/2)?

by jrjack
Tags: eln12 or 2, function
 P: 107 1. The problem statement, all variables and given/known data The full problem is y=e^(x/2) and y=e^(-x) revolved around the x-axis, from [ln6,ln12]. The part I'm struggling with is evaluating 2e^(ln12/2) and 2e^(ln6/2). I am just reviewing for my Calculus 1 final. 2. Relevant equations 3. The attempt at a solution $$2e^{(ln12/2)}$$ $$2 e^{(1/2)(ln12)}$$ $$2 (e^{1/2}), e^{ln12}=12, and 2(12)e^{1/2}???$$
 HW Helper P: 6,191 Well you can easily show that e^(lnA) = A. So you can use that result as is.
 P: 107 The problem is $$e^{(\frac{ln12}{2})}$$ not $$e^{ln\frac{12}{2}}$$, that would be e^(ln6)=6
P: 107

log function e^(ln12/2)?

I have a full equation that looks like this:
$$\pi((2e^{\frac{ln12}{2}}+e^{-12})-(2e^{\frac{ln6}{2}}+e^{-6}))$$
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P: 7,077
 Quote by jrjack 1. The problem statement, all variables and given/known data The full problem is y=e^(x/2) and y=e^(-x) revolved around the x-axis, from [ln6,ln12]. The part I'm struggling with is evaluating 2e^(ln12/2) and 2e^(ln6/2). I am just reviewing for my Calculus 1 final. ...
I doubt that you have stated the full problem ?

What are you finding? Volume, Area, ... ?

What is being revolved around the x-axis ?
 P: 107 Sorry, maybe I have made a mistake in my prior work. (this may need to be moved to the calculus section) The problem states: Use the washer method to find the volume of the solid when the region bounded by y=e^(x/2) and y=e^(-x), x=ln6, x=ln12, when the region is revolved around the x-axis. So I started with: $$\int_{ln6}^{ln12}\pi(e^{\frac{x}{2}}-e^{-x})dx$$ $$\pi(2e^{x/2}+e^{-x})|_{ln6}^{ln12}$$ $$\pi((2e^{\frac{ln12}{2}}+e^{-ln12})-(2e^{\frac{ln6}{2}}+e^{-ln6}))$$
 HW Helper P: 3,326 $$a^{\frac{1}{2}}=\sqrt{a}$$ and $$\left(a^b\right)^c=a^{bc}$$ using these two rules for indices, you should be able to answer your question.
 P: 107 Thanks, so... $$2e^{\frac{ln12}{2}}=2\sqrt{e^{ln12}}=2\sqrt{12}$$ $$=4\sqrt{3}$$ and $$2\sqrt{e^{ln6}}=2\sqrt{6}$$ giving me: $$\pi((4\sqrt{3}-\frac{1}{12})-(2\sqrt{6}-\frac{1}{6}))$$ $$\pi(\frac{1}{12}+4\sqrt{3}-2\sqrt{6})$$ But that was wrong, and I am unsure how to get the correct answer. How does that become $\frac{575\pi}{96}$?
P: 60
 Quote by jrjack giving me: $$\pi((4\sqrt{3}-\frac{1}{12})-(2\sqrt{6}-\frac{1}{6}))$$
No, this isn't true.

$$-\frac{1}{12},-\frac{1}{6}$$ here is mistake.
 Quote by jrjack How does that become $\frac{575\pi}{96}$?
Hm, I checked this in one program, and it isn't true also.
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 Quote by jrjack Sorry, maybe I have made a mistake in my prior work. (this may need to be moved to the calculus section) The problem states: Use the washer method to find the volume of the solid when the region bounded by y=e^(x/2) and y=e^(-x), x=ln6, x=ln12, when the region is revolved around the x-axis. So I started with: $$\int_{ln6}^{ln12}\pi(e^{\frac{x}{2}}-e^{-x})dx$$ ...
$\displaystyle \int_a^b\pi\left((f(x))^2-(g(x))^2\right)\,dx$