<s>[INST] 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?
 [/INST] There are already some great answers to this question, but just in case you need the basics of logs...
One nice way to think about logs is that it asks for the how many of a  particular factor is in a number.
$$log_5(125)=3$$
Because there are three factors of 5 in the number 125.
$$125=5 \cdot 5 \cdot 5=5^3$$
Similarly, 
$$log_3(81)=4$$
Because 
$$81=9 \cdot 9=3 \cdot 3 \cdot 3 \cdot 3=3^4$$
Finding simple logs is just a matter of counting a particular factor.

If you have the log of a fraction like 
$$log_3 \left(\frac{1}{81}\right)$$
Then we are dealing with a negative number of factors.
$$log_3 \left(\frac{1}{81}\right)=-4$$
Because 
$$\frac{1}{81}=3^{-4}$$
Similarly
$$log_2 \frac{1}{8}=-log_2(8)=-3$$

We have one more concept to look at and then we'll be ready.
$$log_{25} (5)$$
We can think of 5 as one half of a factor of 25.
$$log_{25} (5)=\frac{1}{2}$$
Similarly
$$log_{25} (125)=\frac{3}{2}$$
Because there are 3 half factors of 25 in 125.
We can also look at
$$log_{81}(3)=\frac{1}{4}$$
Since 
$$81=3 \cdot 3 \cdot 3 \cdot 3$$
We need four factors of 3 to make 81, so 3 is 1 out of the four factor of 3 we need to make an 81.
$$log_5{\sqrt 5}=\frac{1}{2}$$
Because $\sqrt 5$ is half a factor of 5.
$$log_{25}{\sqrt 5}=\frac{1}{4}$$
Because it takes four factors of $\sqrt{5}$ to make 25.
$$\sqrt 5 \cdot \sqrt 5 \cdot \sqrt 5 \cdot \sqrt 5=25$$
So $\sqrt{5}$ is one fourth of a factor of 25.

So looking at your problem by building up, 
$$log_9(3)=\frac{1}{2}$$
$$log_9(\sqrt{3})=\frac{1}{4}$$
$$log_9\left(\frac{1}{\sqrt{3}}\right)=-\frac{1}{4}$$
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