If $a^x=b$, then $ x=$? Stupid question, I know, but I couldn't remember nor find information by googling on how to find the exponent of $a$ that gives $b$ as the result. If $a^x=b$, then $x=log_a b$ but how do you find $x$? 
 A: For $a,b>0$, take the regular logarithm our of the two sides of the equation.
$a^x=b$ becomes $\ln{a^x}=\ln{b}$
That is $x\ln a=\ln b$
Or $x=\dfrac{\ln b}{\ln a}$
A: Easiest for me to remember is to logarithm both sides (any base log will do) and use the multiplicity:
$$
b = a^x \iff \ln(b) = \ln(a^x) = x \ln(a)
$$
If $\ln(a) \ne 0$ then
$$
 x = \frac{\ln(b)}{\ln(a)}
$$
A: Logarithm function value can't be get in some simple way like sum or product. Any numeric method of finding it is approximate and based on some basic numeric principle.
For example, you may find some $\alpha$ and $\beta$ such that $\alpha < x < \beta$ and make $[\alpha,\beta]$ shorter and shorter until its size becomes smaller than required accuracy. The other way is to use Taylor series and sum some first values of it (the more values, the more accurate the value is). Any calculators and computer programs use algorithms like these.
An old-style and less accurate way is to convert your logarithm to $\ln(b)/\ln(a)$ and then search for $\ln(a')$ and $\ln(b')$ in a logarithmic table, where $a'$ and $b'$ are close enough to $a$ and $b$.
A: $ x = log_a b = \dfrac{\log_u b} {\log_ua} $ where u can be chosen as any real number base, e,10 etc. Each log can be found by looking up a table of logarithms of that base or by using some CAS.
