Solve for $x$ in $\log_4{(x+4)} \le \log_2{(2x+5)}$ 
I would like to solve  $$\log_4{(x+4)} \le \log_2{(2x+5)}$$ for $x$.

I did:
$$\log_4{(x+4)} \le \log_2{(2x+5)} \Leftrightarrow \frac{\log(x+4)}{\log(4)} \le \frac{\log(2x+5)}{\log(2)}\\ \Leftrightarrow \frac{\log(x+4)}{2\log(2)} \le \frac{\log(2x+5)}{\log(2)} \Leftrightarrow \frac{1}{2} \cdot \frac{\log(x+4)}{\log(2)} \le \frac{\log(2x+5)}{\log(2)} \Leftrightarrow\ ???$$
What do I do next?

Hint: Use that $\log(2)>0$ and that $t\mapsto \log(t)$ and $t\mapsto e^t$ are monotonic functions to continue:
\begin{align}
\frac{1}{2} \cdot \frac{\log(x+4)}{\log(2)} \le \frac{\log(2x+5)}{\log(2)} &\iff \frac{1}{2}\log(x+4) \le \log(2x+5) \\ &\iff \log(\sqrt{x+4}) \le \log(2x+5) \\ &\iff \sqrt{x+4} \leq 2x+5\\ &\iff \ldots\end{align}
