Solve $\log_3(3x + 2) = \log_9(4x + 5)$ for $x$ 
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 :D
EDIT: You all solved the first equation, but I still don't understand how to solve the second one (the one solved by WolframAlpha).

Notice, formula $$\color{blue}{log_{b^n}(a)=\frac{1}{n}log_{b}(a)}$$ Now, we have $$log_{3}(3x+2)=log_{9}(4x+5)$$ $$\implies log_{3}(3x+2)=log_{3^2}(4x+5)$$
$$\implies log_{3}(3x+2)=\frac{1}{2}log_{3}(4x+5)$$ $$\implies 2 log_{3}(3x+2)=log_{3}(4x+5)$$  $$\implies log_{3}(3x+2)^2=log_{3}(4x+5)$$ $$\implies (3x+2)^2=4x+5$$ $$\implies 9x^2+12x+4=4x+5$$ $$\implies 9x^2+8x-1=0$$ $$\implies (9x-1)(x+1)=0$$ $$\implies x=\frac{1}{9} \ \text{&} \ x=-1$$ Now, substituitng $x=-1$ , we get $LHS=\log(3(-1)+2)=\log(-1)$ 
But log is not defined for negative number hence the correct solution is $\color{blue}{x=\frac{1}{9}}$
