Solve for x by J. Sacramento

$\begin{aligned} x + \log_{10} (1+2^x) & = x \log_{10} 5 + \log_{10} 6 \\ \log_{10} 10^x + \log_{10} (1+2^x) & = \log_{10} 5^x + \log_{10} 6 & \small \color{#3D99F6} \text{Note: } a\log b = \log b^a \\ \log_{10} \left(10^x (1+2^x)\right) & = \log_{10} \left(6(5^x)\right) & \small \color{#3D99F6} \text{Note: } \log a + \log b = \log (ab) \\ 10^x (1+2^x) & = 6(5^x) & \small \color{#3D99F6} \text{Divide both sides by }5^x \\ 2^x (1+2^x) & = 6 \\ (2^x)^2 + 2^x - 6 & = 0 & \small \color{#3D99F6} \text{A quadratic equation of }2^x. \\ (2^x + 3)(2^x-2) & = 0 \\ 2^x - 2 & = 0 & \small \color{#3D99F6} 2^x+3=0 \text{ has no real solution.} \\ \implies x & = \boxed{1} \end{aligned}$