Evaluate $\log_\frac{1}{3}9$.
I know that the logarithm of a number in a certain base is the exponent that I have to raise the base to in order to get that number. So, I want to find the exponent $x$ that satisfies $\left(\frac{1}{3}\right)^x=9$. I can rewrite both sides of this equation using powers of 3, since $\frac{1}{3}=3^{-1}$ and $9=3^2$. So I get $3^{-x}=3^2$. Since the bases are the same, I can equate the exponents and get $-x=2$. Solving for x, I get $x=-2$. Therefore, $\log_\frac{1}{3}9=-2$. 