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

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?

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 $\color{blue}{x=\frac{1}{9}}$

$$\log_3(3x + 2) = \log_9(4x + 5)$$ Convert the RHS to base $3$ to get $$\log_3 (3x+2) = \frac{\log_3 (4x+5)}{\log_3 3^2}$$

So that you get $$2\log_3 (3x+2) = \log_3 (4x+5)$$

The power law for logarithms yields $$\log_3 (3x+2)^2 = \log_3 (4x+5)$$

Now, "cancelling out the logarithms" gives you $$\bbox[10px, border:solid blue 1px]{(3x+2)^2 = 4x+5}$$ which is an easy quadratic to solve.