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Log function e^(ln12/2)?

by jrjack
Tags: eln12 or 2, function
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jrjack
#1
Dec15-11, 11:05 AM
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

[tex]2e^{(ln12/2)}[/tex]

[tex]2 e^{(1/2)(ln12)}[/tex]

[tex]2 (e^{1/2}), e^{ln12}=12, and 2(12)e^{1/2}???[/tex]
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rock.freak667
#2
Dec15-11, 11:17 AM
HW Helper
P: 6,202
Well you can easily show that e^(lnA) = A. So you can use that result as is.
jrjack
#3
Dec15-11, 11:20 AM
P: 107
The problem is [tex]e^{(\frac{ln12}{2})}[/tex]
not
[tex]e^{ln\frac{12}{2}}[/tex], that would be e^(ln6)=6

jrjack
#4
Dec15-11, 11:24 AM
P: 107
Log function e^(ln12/2)?

I have a full equation that looks like this:
[tex]\pi((2e^{\frac{ln12}{2}}+e^{-12})-(2e^{\frac{ln6}{2}}+e^{-6}))[/tex]
SammyS
#5
Dec15-11, 01:21 PM
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Quote Quote by jrjack View Post
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 ?
jrjack
#6
Dec15-11, 01:35 PM
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:
[tex]\int_{ln6}^{ln12}\pi(e^{\frac{x}{2}}-e^{-x})dx[/tex]

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

[tex]\pi((2e^{\frac{ln12}{2}}+e^{-ln12})-(2e^{\frac{ln6}{2}}+e^{-ln6}))[/tex]
Mentallic
#7
Dec16-11, 08:41 AM
HW Helper
P: 3,531
[tex]a^{\frac{1}{2}}=\sqrt{a}[/tex]

and

[tex]\left(a^b\right)^c=a^{bc}[/tex]

using these two rules for indices, you should be able to answer your question.
jrjack
#8
Dec16-11, 10:58 AM
P: 107
Thanks, so...
[tex]2e^{\frac{ln12}{2}}=2\sqrt{e^{ln12}}=2\sqrt{12}[/tex]
[tex]=4\sqrt{3}[/tex] and
[tex]2\sqrt{e^{ln6}}=2\sqrt{6}[/tex]
giving me:
[tex]\pi((4\sqrt{3}-\frac{1}{12})-(2\sqrt{6}-\frac{1}{6}))[/tex]
[tex]\pi(\frac{1}{12}+4\sqrt{3}-2\sqrt{6})[/tex]
But that was wrong, and I am unsure how to get the correct answer.
How does that become [itex]\frac{575\pi}{96}[/itex]?
Karamata
#9
Dec16-11, 03:34 PM
P: 60
Quote Quote by jrjack View Post
giving me:
[tex]\pi((4\sqrt{3}-\frac{1}{12})-(2\sqrt{6}-\frac{1}{6}))[/tex]
No, this isn't true.

[tex]-\frac{1}{12},-\frac{1}{6}[/tex] here is mistake.
Quote Quote by jrjack View Post
How does that become [itex]\frac{575\pi}{96}[/itex]?
Hm, I checked this in one program, and it isn't true also.
SammyS
#10
Dec16-11, 04:48 PM
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Quote Quote by jrjack View Post
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:
[tex]\int_{ln6}^{ln12}\pi(e^{\frac{x}{2}}-e^{-x})dx[/tex]
...
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:
[itex]\displaystyle \int_a^b\pi\left((f(x))^2-(g(x))^2\right)\,dx[/itex]
You failed to square the two functions.
jrjack
#11
Dec16-11, 05:21 PM
P: 107
Man do I feel like an idiot. Thanks for pointing that out.


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