# Math Help - Log question- need help

1. ## Log question- need help

Hey guys.. I have a log question that I am working on for something I need to hand in today..

I have the answer, I was helped by someone, I am just having trouble understanding a step in the answer, and I don't really want to just move on without being able to grasp whats happening.

Here is the question:

Image of Logs question - Photobucket - Video and Image Hosting

I felt this would be faster then to write it out..
I am wondering what is happening in the 3rd to 4th step mainly.. how does the -1 exponent become -3/2, and what happened to the square root sign and how did that interact?

Thanks alot

2. Note that square roots can be written using fractional exponents: $\sqrt{a} = a^{\frac{1}{2}}$

More generally: $\sqrt[n]{a^m} = \left(\sqrt[n]{a}\right)^m = a^{\frac{m}{n}}$ (if there is no $n$ written, it is implied to be 2)

Here, imagine: $\color{red} a = \tfrac{2}{3}$, $\color{magenta} n = 2$ and $\color{blue} m = 3$

So:

\begin{aligned} 2 + \log_{\frac{2}{3}} \left( {\color{magenta}\sqrt{{\color{black}\bigg( }{\color{red}\frac{2}{3}}{\color{black}\bigg)^{{\c olor{blue}3}}}}}\right)^{-1} & = 2 + \log_{\frac{2}{3}} \left[\left(\frac{2}{3}\right)^{\frac{3}{2}}\right]^{-1} \\ & = 2 + \log_{\frac{2}{3}}\left(\frac{2}{3}\right)^{-1} \qquad \text{Since: } \left(a^b\right)^c = a^{bc} \\ & \ \ \vdots \end{aligned}

3. Hello, Slipery!

$\log_{\frac{2}{3}}\!\left(\frac{4}{9}\cdot\sqrt{\f rac{27}{8}}\right)$
I would do it like this . . .

$\log_{\frac{2}{3}}\! \left[\frac{2^2}{3^2}\cdot\left(\frac{3^3}{2^3}\right)^{ \frac{1}{2}} \right] \;\;=\;\;\log_{\frac{2}{3}}\!\left(\frac{2^2}{3^2} \cdot\frac{3^{\frac{3}{2}}}{2^{\frac{3}{2}}}\right )$ . $= \;\;\log_{\frac{2}{3}}\!\left(\frac{2^{\frac{1}{2} }}{3^{\frac{1}{2}}}\right) \;\;=\;\;\log_{\frac{2}{3}}\!\left(\frac{2}{3}\rig ht)^{\frac{1}{2}}$

. . . $=\;\;\frac{1}{2}\cdot\underbrace{\log_{\frac{2}{3} }\!\left(\frac{2}{3}\right)}_{\text{This is 1}} \;\;=\;\;\frac{1}{2}\cdot1 \;\;=\;\;\frac{1}{2}$

4. You were both very helpful! kind of nice seeing two ways it can be done.

Thank you so much