Solution for $x$ with exponents? I am trying to solve the following,
$$7^{(2x+1)} + (2(3)^x) - 56 = 0$$
Should I put the 56 on the other side and get the log of both sides and is there a better way to solve this.

\begin{align*}
7^{2x+1}+2\cdot3^x-56 = 0 &\iff 7^{2x}\cdot7+2\cdot 3^x-7\cdot 8=0\\ 
&\iff 7(7^{2x}-8)+2\cdot3^x=0 \\
&\iff 7(7^{2x}-8)={-2\cdot3^x}\bf<0
\end{align*}
This is possible if and only if $7^{2x}-8<0$  which is equivalent to $x<\dfrac{\log_78}{2}\approx\dfrac{\log_77}2=0.5$. So we can only hope to find a solution using approximation methods. The answer given by Claude shows that $x$ is pretty close to $0.5$, so we can say: $$\color{grey}{\boxed{\color{white}{\underline{\overline{\color{black}{\displaystyle\, x\approx \dfrac{\log_78}{2}\,}}}}}}$$
