Find all $x$ such that $8^x(3x+1)=4$

Question

Find all $x$ such that $8^x(3x+1)=4$,and prove that you have found all values of $x$ that satisfy this equation.

My effort

Rewriting the equation I have \begin{array} 22^{3x}(3x+1)&=2^2 \\ \log_2{2^{3x}(3x+1)} &=\log_2{2^2} \\ 3x +\log_2{(3x+1)} &=2 \\ \end{array}

I don't know how to simplify from there,looking it up with Geogebra it appears clear that I have only one solution at $x=1/3$ but how can I show this from the last equations ?

If you can give me some hint ,that would be best.

Answer 1

It is clear that any solution must have $3x + 1 > 0$, or $x > -1/3$. For $x > -1/3$, the function $f(x) = 8^x (3x + 1)$ is increasing, hence the equation $f(x) = 4$ has at most one solution. But $x = 1/3$ is a solution, so it is the only one.

Answer 2

You ended up with the system $\log (1+a)=2-a$ where $a=3x$. Observe that $2-a$ is strictly decreasing and $\log (1+a)$ is strictly increasing function of $a$. So there is a unique solution thanks to intermediate value theorem.