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
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"
This is where I get stuck.
Here we go:
$\displaystyle 3x^2=8$
$\displaystyle x^2=\frac{8}{3}$
$\displaystyle x=\sqrt{\frac{8}{3}}$
$\displaystyle x=\frac{\sqrt{8}}{\sqrt{3}}$
$\displaystyle x=\frac{2\sqrt{2}}{\sqrt{3}}$
"Rationalize" the denominator by multipling numerator and denominator by $\displaystyle \sqrt{3}$
$\displaystyle x=\frac{2\sqrt{2}}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}}=\frac{2\sqrt{6}}{3}$
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?