Solve the equation $3^{3x-1}+27^x=2^{2x+1}+7\cdot4^x$ Solve the equation $$3^{3x-1}+27^x=2^{2x+1}+7\cdot4^x$$
We can rewrite the equation as $$\dfrac{3^{3x}}{3}+3^{3x}=2\cdot2^{2x}+7\cdot2^{2x}\\4\cdot3^{3x}=27\cdot2^{2x}\\\dfrac{3^{3x}}{2^{2x}}=\dfrac{27}{4}$$ I don't know how to approach it from here, as the LHS isn't for example $\dfrac{3^{2x}}{2^{2x}}=\left(\dfrac{3}{2}\right)^{2x}$.
The given answer is $x=1$.

$3^{3x-1} + 27^{x} = 2^{2x+1} + 7\cdot4^{x}
\\ 3^{3x}\left(\frac{1}{3} + 1\right) = 2^{2x}(2+7)
\\ 3^{3x}\left(\frac{4}{3}\right) = 2^{2x}\cdot9
\\ \frac{3^{3x}}{2^{2x}} = \frac{27}{4}
\\ \left(\frac{27}{4}\right)^{x} = \frac{27}{4} \\ \textsf{clearly it's x=1}$
