# Question #fce3a

Apr 2, 2017

I am interpreting the question to mean that you want to find ${\log}_{9} \left(\sqrt{\sqrt{\sqrt{3}}}\right)$.

The answer to this is $\frac{1}{16}$.

#### Explanation:

We can expand ${\log}_{9} \left(\sqrt{\sqrt{\sqrt{3}}}\right)$ to ${\log}_{9} \left({\left({\left({3}^{\frac{1}{2}}\right)}^{\frac{1}{2}}\right)}^{\frac{1}{2}}\right)$ (since the square root is equivalent to taking the power of one half).

Now, ${\left({a}^{b}\right)}^{c} = {a}^{b c}$. Using this, we simplify ${\log}_{9} \left({\left({\left({3}^{\frac{1}{2}}\right)}^{\frac{1}{2}}\right)}^{\frac{1}{2}}\right)$ to ${\log}_{9} \left({3}^{\frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2}}\right) = {\log}_{9} \left({3}^{\frac{1}{8}}\right)$.

From here, we just need to remember the logarithm identity $\log \left({a}^{b}\right) = b \log \left(a\right)$. Using this identity, we simplify ${\log}_{9} \left({3}^{\frac{1}{8}}\right)$ to ${\log}_{9} \frac{3}{8}$.

But what is ${\log}_{9} \left(3\right)$? Remember that taking the ${\log}_{b} \left(a\right)$ is finding a number $n$ so that ${b}^{n} = a$. We know that $\sqrt{9} = {9}^{\frac{1}{2}} = 3$. So, ${\log}_{9} \left(3\right) = \frac{1}{2}$.

Then, ${\log}_{9} \frac{3}{8} = \frac{\frac{1}{2}}{8} = \frac{1}{16}$.