# Log function e^(ln12/2)?

## Homework Statement

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.

## The Attempt at a Solution

$$2e^{(ln12/2)}$$

$$2 e^{(1/2)(ln12)}$$

$$2 (e^{1/2}), e^{ln12}=12, and 2(12)e^{1/2}???$$

rock.freak667
Homework Helper
Well you can easily show that e^(lnA) = A. So you can use that result as is.

The problem is $$e^{(\frac{ln12}{2})}$$
not
$$e^{ln\frac{12}{2}}$$, that would be e^(ln6)=6

I have a full equation that looks like this:
$$\pi((2e^{\frac{ln12}{2}}+e^{-12})-(2e^{\frac{ln6}{2}}+e^{-6}))$$

SammyS
Staff Emeritus
Homework Helper
Gold Member

## Homework Statement

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 ?

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:
$$\int_{ln6}^{ln12}\pi(e^{\frac{x}{2}}-e^{-x})dx$$

$$\pi(2e^{x/2}+e^{-x})|_{ln6}^{ln12}$$

$$\pi((2e^{\frac{ln12}{2}}+e^{-ln12})-(2e^{\frac{ln6}{2}}+e^{-ln6}))$$

Last edited:
Mentallic
Homework Helper
$$a^{\frac{1}{2}}=\sqrt{a}$$

and

$$\left(a^b\right)^c=a^{bc}$$

using these two rules for indices, you should be able to answer your question.

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 $\frac{575\pi}{96}$?

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.
How does that become $\frac{575\pi}{96}$?
Hm, I checked this in one program, and it isn't true also.

Last edited:
SammyS
Staff Emeritus
Homework Helper
Gold Member
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:
$$\int_{ln6}^{ln12}\pi(e^{\frac{x}{2}}-e^{-x})dx$$
...
Using the washer method to find the volume of the solid when the region bounded by y=f(x) and y=g(x), x=a, x=b, is revolved around the x-axis, (Assumes f(x)>g(x) on the interval [a,b].) results in the following integral:
$\displaystyle \int_a^b\pi\left((f(x))^2-(g(x))^2\right)\,dx$​

You failed to square the two functions.

Man do I feel like an idiot. Thanks for pointing that out.