Solve the equation $4^{7 x - 10} = 10^{7 x - 6}$ for $x$ Solve for $x$ of the following equation. 
$$4^{7 x - 10} = 10^{7 x - 6}$$
I tried to make $4$ and $10$ have a common base but I could not find one so I don't know where to go from here. 
 A: $$
\begin{align}
& 4^{7 x - 10} = 10^{7 x - 6} \\[6pt]
\implies & \log_4(4^{7 x - 10}) = \log_4(10^{7 x - 6}) \\[6pt]
\implies & 7x-10 = (7x-6)\log_4(10) \\[6pt]
\implies & 7x-7x\log_4(10) = 10-6\log_4(10) \\[6pt]
\implies & 7x[1-\log_4(10)] = 10-6\log_4(10) \\[6pt]
\implies & 7x = \frac{10-6\log_4(10)}{1-\log_4(10)} \\[6pt]
\implies & x = \frac{10-6\log_4(10)}{7-7\log_4(10)}
\end{align}
$$
A: Hint
$$4^{7x-10}=10^{7x-6}=10^4 10^{7x-10}\implies\left(\frac{2}{5}\right)^{7x-10} =\left(\frac{4}{10}\right)^{7x-10}=10^4.$$
Now take logarithms and you will get an equation in $x.$
A: You don't need both sides to have the same base.  You can take $\ln$ of both sides.  This allows you to bring the exponents down.  
If you do want both sides to have the same base then convert each side to base $e$, note that for any number $a > 0$ and $a \neq 1$ we have that 
$a^b = e^{b \ln(a)}$
Rewriting both sides to have base $e$ and then equating exponents is the same thing as taking $\ln$ of both sides.
A: $$4^{7x - 10} = 10^{7x - 6}$$
$$(7x - 10)\log4 = (7x - 6)\log10$$
$$7x = \frac{6\log10 - 10\log4}{\log10 - \log4}$$
$$ = \frac{6\log5 + 6\log2 - 20\log2}{\log5 + \log2 - 2\log2}$$
$$ = \frac{6\log5 - 14\log2}{\log5 - \log2}$$
$$ x = \frac{2(3\log5 - 7\log2)}{7(\log5 - \log2)}$$
