Exponent equation with common power Solve for $x$ in
$$\log_{2}(2^{x-1}+3^{x+1}) = 2x-\log_{2}(3^x)$$
I simplified by doing:
$$\log_{2}(3^x \cdot 2^{x-1} + 3^{2x+1}) = 2x$$
$$\frac{6^x}{2} + 3^{2x+1} = 2^{2x}$$
$$6^x + 2 \cdot 3^{2x+1} = 2^{2x+1}$$
Where do I go from here? I tried moving more numbers around, but I haven't been able to get any closer to solving for x. Any help is appreciated.

$$6^x + 2 \cdot 3^{2x+1} = 2^{2x+1}$$
$$2^x\cdot 3^x+2 \cdot 3^{2x+1} - 2^{2x+1}=0$$
divide by $2^x\cdot 3^x$
$$1+\frac{2 \cdot 3^{2x+1}}{2^x\cdot 3^x}-\frac{2^{2x+1}}{2^x\cdot 3^x}=0$$
$$1+6 \cdot\frac{ 3^{x}}{2^x}-2\cdot\frac{2^{x}}{ 3^x}=0$$
Set $\left(\frac23\right)^x=z$
$$1+\frac{6}{z}-2z=0$$
$$2z^2-z-6=0\to z_1=2;\;z_2=-\frac{3}{2}$$
$$\left(\frac23\right)^x=2\to x\ln \frac23=\ln 2\to x=\frac{\ln  2}{\ln 2 -\ln 3}$$
$\left(\frac23\right)^x=-\frac{3}{2}$ has no real solutions
