How to find $x$ given $\log_{9}\left(\frac{1}{\sqrt3}\right) =x$ without a calculator? I was asked to find $x$ when: $$\log_{9}\left(\frac{1}{\sqrt3}\right) =x$$ 
Step two may resemble: 
$${3}^{2x}=\frac{1}{\sqrt3}$$
I was not allowed a calculator and was told that it was possible. I put it into my calculator and found out that $x$=-0.25 but how do you get that?

$\log_{9}\left(\frac{1}{\sqrt3}\right) = $
$\log_{9}(3^{-\frac 12})=$
$\log_{9}((\sqrt{9})^{-\frac 12})=$
$\log_{9}((9^{\frac 12})^{-\frac 12}) = $
$\log_{9}(9^{\frac 12*(-\frac 12}) =$
$\log_9(9^{-\frac 14}) =$
$-\frac 14$
..... or .....
$\log_9(\frac 1{\sqrt 3}) =x$ So 
$9^x = \frac 1{\sqrt 3}= \frac 1{\sqrt{\sqrt 9}} = \frac 1{\sqrt[4]{9}} = 9^{-\frac 14}$
$x = -\frac 14$
