Solve for $x$: $\log_3(x-2)\ge\log_5(4-x)$

Question

Solve for $x$: $\log_3(x-2)\ge\log_5(4-x)$

$x-2\gt0\implies x\gt2$

$4-x\lt0\implies x\lt4$

$$\frac{\log(x-2)}{\log3}-\frac{\log(4-x)}{\log5}\ge0$$

$$\implies\log5\log(x-2)-\log3\log(4-x)\ge0$$

$$\implies\log\frac{(x-2)^{\log5}}{(4-x)^{\log3}}\ge0$$

$$\implies\frac{(x-2)^{\log5}}{(4-x)^{\log3}}\ge1$$

$$\implies\frac{(x-2)^{\log5}-(4-x)^{\log3}}{(4-x)^{\log3}}\ge0$$

Not able to proceed next. Also, not sure if my approach is correct.

Answer 1

$\log_3(x-2)=\log_5(4-x)$ when $x=3$.
Note that $\log_3(x-2)$ is a increasing function while $\log_5(4-x)$ is a decreasing function.
So $3\leq x<4$

Answer 2

$$ \log_3 (x-2) \ge \log_5(4-x) \\ \frac{\ln (x-2)}{\ln 3} \ge \frac{\ln (4-x)}{\ln 5} $$ Suppose $\ln(4-x) \gt 0 \iff x\lt 3$. Then $$\frac{\ln (x-2)}{\ln(4-x)} \ge \frac{\ln 3}{\ln 5}\\ \log_{4-x} (x-2) \ge \log_5 3 \\ $$ But this cannot be true as the LHS is negative as $x-2 \lt 1$, while the RHS is positive. On the other hand, if $\ln(4-x) \le 0 \iff 3\le x\lt 4$, then $$\log_{4-x} (x-2) \le \log_53 $$ This is always true as the LHS is non-positive ($\log_a b \le 0$ if $a\lt 1$ and $b\ge 1$) while the RHS is positive.

So, the solution set is $[3,4)$.