Solution

Evaluate the following:

$$\log_4 8$$


The first thing that should cross one's mind is that $\log_b x$ is the "exponent one needs on $b$ to produce $x$". This is often enough to solve the problem if the exponent is a small integer. For example $\log_2 8 = 3$, as $2^3 = 8$. Likewise $\log_2 4 = 2$ as $2^2 = 4$.

Since $8$ is between the powers $4^1$ and $4^2$, we know the exponent we seek is a fraction, so things will be a little bit harder.

With a little bit of insight, we might notice that both $4$ and $8$ are "nice powers" of $2$. So, let's see if we can't use $2$ as a "stepping stone" to get from $4$ to $8$...

If one takes a square root of $4$, one gets $2$. Then if one cubes $2$ one gets $8$.

Thus, $8 = (4^{1/2})^3 = 4^{3/2}$ and consequently $\log_4 8 = 3/2$.




Alternate Solution

Noting that both $4$ and $8$ are both "nice" powers of $2$ -- we could appeal to the change-of-bases theorem $$\log_4 8 = \frac{\log_2 8}{\log_2 4} = \frac{3}{2}$$


Another Alternate Solution

Suppose we let $y = \log_4 8$ and then convert things to exponential form first...

$$\begin{array}{rcll} 4^y &=& 8\\ (2^2)^y &=& 2^3 & \scriptsize{\textrm{both } 4 \textrm{ and } 8 \textrm{ are representable as powers of } 2, \textrm{ so we use this as a common base}}\\ 2^{2y} &=& 2^3 & \scriptsize{\textrm{recall if } b^m = b^m, \textrm{ then it must be true that } m = n, \textrm{ so...}}\\ 2y &=& 3 & \\ y &=& \frac{3}{2} \end{array}$$