[Math] Solve $\log_3(3x + 2) = \log_9(4x + 5)$ for $x$

Question

Solve for $x$

$$
\log_3(3x + 2) = \log_9(4x + 5)
$$

I changed the bases of the logs

$$
\frac {\log_{10}(3x + 2)} {\log_{10}(3)} = \frac {\log_{10}(4x + 5)} {\log_{10}(9)}
$$

Now I'm stuck, I don't know how to eliminate the logs.

On WolframAlpha I've seen that $\dfrac {\log_{10}(3x + 2)} {\log_{10}(3)} = 0$ gives $ x = -\frac{1}{3}$.
Do you know how does this equation and the equation above can be solved?

Thanks in advance, first question here 😀

EDIT: You all solved the first equation, but I still don't understand how to solve the second one (the one solved by WolframAlpha).

Answer 1

All you need is $\log_9 = \frac12\log_3$: $$ \log_3 (3x+2) = \frac12\log_3(4x+5)\Longrightarrow (3x+2)^2 = 4x+5 $$ Could you proceed?