Solution for $x$ with exponents?

Question

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.

Answer 1

\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: $$\color{grey}{\boxed{\color{white}{\underline{\overline{\color{black}{\displaystyle\, x\approx \dfrac{\log_78}{2}\,}}}}}}$$

Answer 2

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.

Edit (ten years later !)

It is better to consider $$g(x)=\log \left(2\, 3^x+7^{2 x+1}\right)-\log (56)=0$$ which is very close to linearity.

Using $x_0=\frac 12$, the first iterate of Newton method is $$x_1=\frac 12+\frac{\left(49+2 \sqrt{3}\right) \log \left(\frac{56}{49+2 \sqrt{3}}\right)}{98 \log (7)+\sqrt{3} \log (9)}=0.517593$$

Answer 3

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