Evaluate the following:

$$\log_{1/9} \sqrt{27}$$

The first thing that should cross one's mind is that $\log_b x$ is the "exponent one needs on $b$ to produce $x$".

Noticing that both $1/9$ and $27$ are "nice powers" of $3$, let's see if we can use $3$ as a "stepping stone" to get from our base of $1/9$ to $\sqrt{27}$...

If one recipricates $1/9$ and then takes a square root of the result, one gets $3$. Then if one cubes $3$ and takes a square root of the result, one gets $\sqrt{27}$.

Remembering that reciprocation is the same as raising something to the $-1$ power and taking a square root is equivalent to raising something to the $1/2$ power, we have $$\sqrt{27} = (((3^{-1})^{1/2})^3)^{1/2} = 3^{-3/4}$$


$$\log_{1/9} \sqrt{27} = -\frac{3}{4}$$

Alternate Solution

We can use the properties of logarithms to simplify the logarithms present down to logarithms that can be evaluated by inspection...

$$\begin{array}{rcll} \log_{1/9} \sqrt{27} &=& \frac{\log_3 \sqrt{27}}{\log_3 \frac{1}{9}} & \scriptsize{\textrm{noting both } 9 \textrm{ and } 27 \textrm{ are powers of } 3 \textrm{ we start with the change of bases theorem...}}\\\\ &=& \frac{\log_3 27^{1/2}}{\log_3 9^{-1}} & \scriptsize{\textrm{recalling square roots and recipricals can be expressed in terms of exponents}}\\\\ &=& \frac{\frac{1}{2} \log_3 27}{-\log_3 9} & \scriptsize{\textrm{recalling } \log_b x^n = n\log_b x}\\\\ &=& \frac{\frac{1}{2} \cdot 3}{-2} & \scriptsize{\textrm{both logarithms above were of integer powers of } 3 \textrm{ so they were easy to evaluate}}\\\\ &=& -\frac{3}{4} &\end{array}$$

Another Alternate Solution

Suppose we let $y = \log_{1/9} \sqrt{27}$ and then convert things to exponential form first...

$$\begin{array}{rcll} \left( \frac{1}{9} \right)^y &=& 27^{1/2}\\\\ (3^{-2})^y &=& (3^3)^{1/2} & \scriptsize{\textrm{both } 1/9 \textrm{ and } \sqrt{27} \textrm{ are representable as powers of } 3, \textrm{ so we use this as a common base}}\\\\ 3^{-2y} &=& 3^{3/2} & \scriptsize{\textrm{recall if } b^m = b^m, \textrm{ then it must be true that } m = n, \textrm{ so...}}\\\\ -2y &=& \frac{3}{2} & \\\\ y &=& -\frac{3}{4} \end{array}$$