How do you solve a logarithm with a non-integer base? How would one calculate the log of a number where the base isn't an integer (in particular, an irrational number)? For example:
$$0.5^x = 8 \textrm{ (where } x = -3\textrm{)}$$
$$\log_{0.5}8 = -3$$
How would you solve this, and how would this work for an irrational base (like $\sqrt{2}$)?
 A: This is the base changing formula : 
$$\log_{a}(x) = \frac{\log_{b}(x)}{\log_{b}(a)}$$
A: rewrite your equation in the form $2^{-x}=2^{3}$
A: Let's rewrite this in a different way:
$$
0.5^x=8
$$
Take the logarithm with respect to any base $a$ ($a>0$, $a\ne1$):
$$
\log_a(0.5^x)=\log_a8
$$
which becomes
$$
x\log_a 0.5=\log_a 8
$$
or
$$
x=\frac{\log_a 8}{\log_a 0.5}
$$
You would stop here weren't from the fact that $8=2^3$ and $0.5=2^{-1}$, so
$$
x=\frac{\log_a 8}{\log_a 0.5}=\frac{\log_a 2^3}{\log_a 2^{-1}}=
\frac{3\log_a 2}{-\log_a 2}=-3
$$
You need to compute no logarithm, actually.
A: Yet another attempt to help. Let’s look at a somewhat different case. How do you calculate $\log_{3.8}(8)$? Again as egreg says, rewrite as $3.8^x=8$, take the (ordinary) log of both sides to get $x\log_{10}(3.8)=\log_{10}(8)$, and $x=\log_{10}(8)/\log_{10}(3.8)$
If you want $\log_{\sqrt2}(8)$, it’s the same thing, only you can shortcut matters early: $8={\sqrt2\,}^x=(2^{1/2})^x=2^{x/2}$; knowing that $8=2^3$, you see that $x=6$, $\log_{\sqrt2}(8)=6$.
A: You can use natural logarithms, because for all $a$ and $b$ : $\log_a(b)=\frac{\ln(b)}{\ln(a)}$.
