Solve for $x$ in $\log_4{(x+4)} \le \log_2{(2x+5)}$

Question

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?

Answer 1

Since the following identity holds for all $u>0$: $$ \log_{2}\left(u\right) = 2 \log_{4}\left(u\right) = \log_{4}\left(u^2\right) $$ The original inequality $\log_{4}\left(x+4\right) \leqslant \log_{2}\left(2 x + 5\right)$ can be rewritten as $$\log_{4}\left(x+4\right) \leqslant \log_{4}\left(\left(2 x + 5\right)^2 \right)$$ Since $\log_{4}\left(u\right)$ is an increasing function for $u>0$, this implies $$ x + 4 \leqslant \left(2 x+5\right)^2 $$ which has to be supplemented with auxiliary conditions $x+4 > 0$ and $2 x+5 > 0$. This results in the answer $$ x \geqslant - \frac{7}{4} $$

Answer 2

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}

Answer 3

Remove the common denominator $\log 2$ and multiply by $2$ to clear the fractions. Thus

$\log(x+4)\leq 2 \log(2x+5)$.

The right side of the above is also $\log((2x+5)^2)$ from the properties of logarithms, so we are reduced to the inequality

$x+4\leq (2x+5)^2$

which can be handled by ordinary algebraic methods (which I leave to the questioner or reader). You must limit the solution set to a subset of $x > (-(5/2))$ because the logarithms (for real arithmetic) are defined only for positive arguments.