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.
Put $3^x=a$ and $2^x=b$. Then
$$6a^2+ab-2b^2=0$$ $$\Rightarrow 6a^2+4ab-3ab-2b^2=0$$ $$\Rightarrow (3a+2b)(2a-b)=0$$
Now $3a+2b$ is always positive. So $2a-b=0$.
Therefore, $$2\cdot 3^x = 2^x$$ $$\Rightarrow x=\log_{2/3} 2=\frac{1}{\log_2 (2/3)}$$
$$6^x + 2 \cdot 3^{2x+1} = 2^{2x+1}$$ $$2^x\cdot 3^x+2 \cdot 3^{2x+1} - 2^{2x+1}=0$$ 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
