Could anyone explain this curious solution to $7^{2x}= 2^x$ I typed the following on wolfram alpha today : $7^{2x} = 2^x$ and found this as a solution besides $x=0$:
$x = \dfrac{2\pi i n}{\log2 - 2\log7}$ where $log$ has a base of $e$ and $n$ is any integer.
I am scratching my head wondering where they got this solution. 
 A: Solutions to $e^z=1$ are $2\pi i\Bbb Z$ by Euler's formula
$\Rightarrow$ solutions to $a^z=1~\,$ ($\Leftrightarrow e^{(\ln a)z}=1$)$~$ are $~\frac{2\pi i}{\ln a}\Bbb Z~\,$ (say $a\in \Bbb R^+$)
$\Rightarrow$ solutions to $a^z=b^z\,$ ($\Leftrightarrow (a/b)^z=1$) are $\frac{2\pi i}{\ln(a/b)}\Bbb Z$ (again $a,b>0$)
A: This is a result of the fact that in the complex numbers, for any integer $n$,
$$
e^{2 \pi i n} = 1
$$
In particular, this means that
when you take a logarithm in the complex numbers, you get multiple possible answers--you have to add some multiple of $2\pi i$.  So to solve the equation, "$a^x = b$" for $x$, when $a$ and $b$ are real, you write
$$
\ln(a^x) = \ln b + 2 \pi i n
$$
$$
x \ln(a) = \ln b + 2 \pi i n
$$
$$
x = \frac{\ln b + 2 \pi i n}{\ln a}
$$
Your equation in particular is $7^{2x} = 2^x$, which is the same as $49^x = 2^x$, which is the same as $(49/2)^x = 1$.  So the above formula gives the answer as
$$
x = \frac{\ln 1 + 2 \pi i n}{\ln(49/2)} = \frac{2 \pi i n}{2 \ln 7 - \ln 2}.
$$
(You will notice the denominator has the opposite sign in your question, but this does not matter, because $n$ ranges over positive and negative integers.)
