Logarithmic function Compute the value of $f\bigl(\frac{1}{400}\bigr)$ if the function is defined as follows :
$f(xy) = f(x) + f(y)$ and $f(4)= 16$ 
 A: To answer this question, we need to assume $f$ is continuous. If $f$ is not continuous, there is no way, with the information given, to get an answer.
The relation
$$
f(xy)=f(x)+f(y)\tag{1}
$$
implies by induction
$$
nf(x)=f(x^n)\quad\text{for }n\in\mathbb{Z}\tag{2}
$$
We can infer from $(2)$ that
$$
f(a^q)=qf(a)\tag{3}
$$
for any $a\gt0$ and $q\in\mathbb{Q}$, and if $f$ is continuous, then $(3)$ holds for all $q\in\mathbb{R}$.
By the definition of $\log$, $(3)$ says
$$
\frac{f(x)}{f(a)}=\log_a(x)\tag{4}
$$

$$
\begin{align}
\frac{f\left(\frac1{400}\right)}{f(4)}
&=\log_4\left(\frac1{400}\right)\\
&=-\log_4(400)\\
&=-\left(2+\log_4(25)\right)\\
&=-\left(2+\log_2(5)\right)\\
f\left(\frac1{400}\right)
&=-16\left(2+\log_2(5)\right)\\
&\doteq-69.1508495181978
\end{align}
$$
A: What happens when $z=1$? Does that teach you something about the behavior of $f$?
Suppose $f$ is indeed a logarithm. Then for some $b$,
$$f(x)=\log_b x = \frac{\ln x}{\ln b}.$$
A little algebraic manipulation will then go a fairly long way ($b$ will go away).
A: The value 2^(1/8) you obtained for the base is correct. But for f(1/400), I am afraid that you did not look properly at the graph or the graph is wrong; the approximate value is -69.15 (its exact value on the basis of natural logarithms is - 8 Log[400] / Log[2]). Please have a look to the formulas related to the change of base for logarithms
(http://www.proofwiki.org/wiki/Change_of_Base_of_Logarithm).
