Giving an Exact Solution to a Logarithm

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• April 10th 2012, 06:17 PM
Mollers
Giving an Exact Solution to a Logarithm
Subject are exactly what the instructions are and I have no problem doing logarithms until I came upon this problem. Teacher didn't really give any examples like this one and I am totally lost.

I have no idea how to write that exactly as I see it
Any help is GREATLY appreciated thank you!Attachment 23557
• April 10th 2012, 07:35 PM
Prove It
Re: Giving an Exact Solution to a Logarithm
\displaystyle \begin{align*} \log_2{\sqrt{3x^2}} &= \frac{3}{2} \\ \sqrt{3x^2} &= 2^{\frac{3}{2}} \end{align*}

Go from here.
• April 10th 2012, 08:16 PM
Mollers
Re: Giving an Exact Solution to a Logarithm
Attachment 23561

Thank you for the reply.
I tried that earlier and just got stuck again as I've never done a logarithm like this.

Am I on the right track? If so I have no idea what to do next =(
• April 10th 2012, 08:19 PM
Mollers
Re: Giving an Exact Solution to a Logarithm
The answer I got is X is approximately .9428
• April 10th 2012, 10:12 PM
browni3141
Re: Giving an Exact Solution to a Logarithm
Quote:

Originally Posted by Mollers
The answer I got is X is approximately .9428

"approximately" isn't exact though is it ;)
This problem is completely solvable without a calculator. No decimals!
• April 10th 2012, 10:27 PM
princeps
Re: Giving an Exact Solution to a Logarithm
$\log_2(\sqrt{3x^2})=\frac{3}{2}$

$\frac{1}{2} \log_2(3x^2)=\frac{3}{2}$

$\log_2(3x^2)=3$

$3x^2=2^3$

$\vdots$
• April 10th 2012, 10:34 PM
biffboy
Re: Giving an Exact Solution to a Logarithm
I'm using the fact that log(sqrty)=1/2logy
So from your question we get 1/2log3x^2=3/2 log3x^2=3 So 3x^2=2^3 (since log base is 2)
So x^2=8/3 x=sqrt(8/3) =root8/root3 =2root2/root3
Multiply top and bottom by root3 Get x=2root6/3
• April 11th 2012, 07:51 AM
Mollers
Re: Giving an Exact Solution to a Logarithm
First off thank you all for the help I couldn't get no where near where I am in this problem without you. I am stuck again as I cannot figure out how to write the last part of what Biffboy said..."root8/root3 =2root2/root3 --------> Multiply top and bottom by root3 Get x=2root6/3"

Attachment 23566

This is where I get stuck.
• April 11th 2012, 07:52 AM
Mollers
Re: Giving an Exact Solution to a Logarithm
ignore the bottom thumbnail
• April 11th 2012, 08:24 AM
masters
Re: Giving an Exact Solution to a Logarithm
Quote:

Originally Posted by Mollers
First off thank you all for the help I couldn't get no where near where I am in this problem without you. I am stuck again as I cannot figure out how to write the last part of what Biffboy said..."root8/root3 =2root2/root3 --------> Multiply top and bottom by root3 Get x=2root6/3"

Attachment 23566

This is where I get stuck.

Here we go:

$3x^2=8$

$x^2=\frac{8}{3}$

$x=\sqrt{\frac{8}{3}}$

$x=\frac{\sqrt{8}}{\sqrt{3}}$

$x=\frac{2\sqrt{2}}{\sqrt{3}}$

"Rationalize" the denominator by multipling numerator and denominator by $\sqrt{3}$

$x=\frac{2\sqrt{2}}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}}=\frac{2\sqrt{6}}{3}$
• April 11th 2012, 09:21 AM
Mollers
Re: Giving an Exact Solution to a Logarithm
thank you so much! I dont understand why I was given this problem when I was never given an example that was anywhere near similar.
• April 12th 2012, 10:31 AM
Mollers
Re: Giving an Exact Solution to a Logarithm
I have been looking at this problem more and thinking about it and wanted to make sure about something.

Since the directions are to give an exact solution would I leave the answer as 2 sqaure root 6/3 or should I compute this and get a decimal answer?
• April 12th 2012, 10:56 AM
biffboy
Re: Giving an Exact Solution to a Logarithm
Leave it as it is. Any decimal answer you got would be an approximation.