How to solve this exponential equation? $2^{2x}3^x=4^{3x+1}$. I haven't been able to find the correct answer to this exponential equation:
$$\eqalign{
2^{2x}3^x&=4^{3x+1}\\
2^{2x} 3^x &=  2^2 \times 2^x \times 3^x\\
4^{3x+1} &= 4^3 \times 4^x \times 4\\
6^x \times 4 &= 4^x \times 256\\
x\log_6 6 + \log_6 4 &= x\log_64 + \log_6 256\\
x + \log_6 4 &= x\log_64 + \log_6 256\\
x-x\log_6 4 &= \log_6 256 - \log_6 4\\
x(1-\log_6 4) &= \log_6 256 - \log_6 4\\
x &= \dfrac{\log_6 256 - \log_6 4}{1-\log_6 4}\\
x &= 10.257}$$
so when I checked the answer I wasn't able to make them equal, I have tried variants of this method but I feel I'm missing something.. 

I think the problem with your solution might be here:
$2^2x * 3^x => 2^2 * 2*x * 3^x$
$4^{3x+1} => 4^3 * 4^x * 4$
since $a^{bc} = (a^b)^c$ and not $a^b * a^c$. So for instance:
$4^{3x+1} => 4^x * 4^x * 4^x * 4$
