<s>[INST] 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.. 
 [/INST] Your second step goes wrong
$2^{2x}=2^x+2^x$ but not $2^2*2^x$
simpler way
$3^x=(2^2)^{3x+1}/2^{2x}$
$3^x=2^{6x+2-2x}$
$3^x=2^{4x+2}$
Then apply log to the base e on both sides and use your calculator.
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