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.
 A: Equation $$f(x)=7^{(2x+1)} + (2(3)^x) - 56 = 0$$ cannot be solved using elemental functions. What you can observe is that $f(0)=-47$ and $f(1)=293$. So, there is a solution for $0<x<1$. If you further refine, you could notice that $f(1/2)=2 \sqrt{3}-7=-3.5359$, so the solution is pretty close to $x=0.5$ (just above since $f(x)$ varies extremely fast). For polishing the root (which is $x=0.517573$), you need to use a method such as Newton or secant.  
If you want, I could elaborate tomorrow morning (I have to go now). Cheers and let me know.
A: For sure this $x$ does not come smoothly (Not integers)....
Consider $7^{(2x+1)} + (2(3)^x) - 56 = 0$ and suppose $x$ is an integer 


*

*As $7$ divides $7^{(2x+1)}$ and $56$ we should have :  $7$ divides $2.3^x$ 

*As $2$ divides $2(3)^x$ and $56$ we should have :  $2$ divides $7^{2(3x+1)}$ 
Neither of this makes sense...
so, your $x$ is not an integer...
It is up to you to see prove that this is not even a rational number...
So, your $x$ is an irrational number and i do not yet know any method other than that of above two methods to go near $x$
This only shows $x$ is an irrational number
A: \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: $$\color{grey}{\boxed{\color{white}{\underline{\overline{\color{black}{\displaystyle\, x\approx \dfrac{\log_78}{2}\,}}}}}}$$
