Maybe I'm completely wrong but I don't get it. I calculate the values of $3^n$ in Excel (in a range from $0$ to $30$). Then I take the values and calculate the logarithm of it but to a different base of $2$. When I plot the result I get a linear curve and it doesn't matter if I take as base $3$ or $2$.
So why is the logarithm to any base from a exponential function to any base always a linear function? Or isn't it? Thanks!
It is.
First, $\log_a (b)$ is defined to be $\frac{\ln b}{\ln a}$, where $\ln$ denotes the natural logarithm. (You then have to prove that $a^{\log_a b} = b$.)
Let's look at $$ \log_a (s^q) $$ so $s$ is the base you use for exponents ($3$ in your example), and $a$ is the base oyu use for logs ($2$ in your example): \begin{align} \log_a (s^q) &= \frac{\ln s^q}{\ln a} & \text{using the definition}\\ & = \frac{1}{\ln a} \ln s^q& \text{algebra}\\ & = \frac{1}{\ln a}q \ln s & \text{property of $\ln$}\\ & = \frac{\ln s}{\ln a} q & \text{algebra} \end{align} which is just a constant multiple of $q$.
As for the proof that's needed to show that definition is correct, you need to know that for natural logs, $\ln (xy) = \ln x + \ln y$ (the "multiplication rule"), and that if $\ln x = \ln y$, then $x = y$ (the "1-1 property"). One consequence of the multiplication rule is that $\ln x/y = \ln x - \ln y ("division rule") and that $\ln x^y = y \ln x$. (Note: all these make sense only when the arguments to $\ln$, on both sides, are positive.)
Let's compare $$ a^{\log_a b} $$ with $b$, by comparing their logarithms. That is, we'll see that $$ \ln(a^{\log_a b}) $$ is the same as $$ \ln b, $$ and thus, but the "1-1" property of logs, see that the two are the same.
OK. Well \begin{align} \ln(a^{\log_a b}) &= \log_a (b) \cdot \ln(a) & \text{by the power rule} \\ &= \frac{\ln(b)}{\ln (a)} \cdot \ln(a) & \text{by the definition of $\log_a$} \\ &= \ln(b) & \text{by cancellation.} \\ \end{align}
So we've shown that these two numbers have the same natural log, hence (by the 1-1 rule) they are the same.