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Log function e^(ln12/2)? 
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#1
Dec1511, 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 xaxis, 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] 


#2
Dec1511, 11:17 AM

HW Helper
P: 6,208

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



#3
Dec1511, 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 


#4
Dec1511, 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] 


#5
Dec1511, 01:21 PM

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P: 7,782

What are you finding? Volume, Area, ... ? What is being revolved around the xaxis ? 


#6
Dec1511, 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 xaxis. 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] 


#7
Dec1611, 08:41 AM

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P: 3,502

[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. 


#8
Dec1611, 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]? 


#9
Dec1611, 03:34 PM

P: 60

[tex]\frac{1}{12},\frac{1}{6}[/tex] here is mistake. 


#10
Dec1611, 04:48 PM

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[itex]\displaystyle \int_a^b\pi\left((f(x))^2(g(x))^2\right)\,dx[/itex]You failed to square the two functions. 


#11
Dec1611, 05:21 PM

P: 107

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



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