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.. 
 A: You made a mistake in your first steps.
It should be :
$$4^{3x+1}= (4^3)^x \cdot 4$$ 
$$2^{2x} 3^x= 4^x 3^x$$
A: 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:
$4^{3x+1} => 4^x * 4^x * 4^x * 4$
A: $$2^{2x}3^x=4^{3x+1},$$
$$(e^{\log2})^{2x}(e^{\log3})^x=(e^{\log4})^{3x+1},$$
$$e^{2x\log2}e^{x\log3}=e^{(3x+1)\log4},$$
$$e^{2x\log2+x\log3}=e^{(3x+1)\log4}.$$
Take the log,
$$2x\log2+x\log3=(3x+1)\log4,$$and
$$x=\frac{\log4}{2\log2+\log3-3\log4}=\frac1{\frac{\log3}{\log4}-2}.$$
A: Take the logarithm : you get
$$ 2x\ln 2+x\ln 3=(3x+1)\ln 4$$
Then solve this linear equation to get the desired result.
Added in edit
First use $\ln4=2\ln2$ and then group the $x$ together on the left-hand side.
$$ x\times(2\ln2+\ln3-6\ln2)=2\ln2.$$
Now simplify and divide
$$ x=\frac{2\ln2}{\ln3-4\ln2}\simeq -0.82814445.$$
A: 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.
