algebra precalculus - 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?

Answer

You are asked to find $\displaystyle \log_{9}\big(\frac{1}{\sqrt3}\big)$, which means a number $x$ such that
$$9^x = \frac{1}{\sqrt3}$$

At this point, it is natural to express everything as powers of $3$. (I admit that for this to come naturally takes some skill and experience: for example, you need to be able to see that $9 = 3^2$ and that $\sqrt{3} = 3^{1/2}$.) So:

$$3^{2x} = 3^{-1/2}$$

At this point it is a simple equation:

$$2x = -\frac12,$$

or $$x = -\frac14.$$