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

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.

$$\log_3(x-2)=\log_5(4-x)$$ when $$x=3$$.
Note that $$\log_3(x-2)$$ is a increasing function while $$\log_5(4-x)$$ is a decreasing function.
So $$3\leq x<4$$
• Since $\frac{(x-2)^{\log5}}{(4-x)^{\log3}} \ge 1$, we need only ask, "for what values of $x$ is the numerator larger?" Since the numerator is equal to the denominator at $x=3$ and the numerator is increasing while the denominator decreases, the range of values for $x$ are $3 \le x <4$. Jun 5, 2021 at 21:47
• @aarbee $\log_3(x-2)>0>\log_5(4-x)$ when $x>3$ and $\log_3(x-2)<0<\log_5(4-x)$ when $x<3$. Jun 6, 2021 at 0:40
$$\log_3 (x-2) \ge \log_5(4-x) \\ \frac{\ln (x-2)}{\ln 3} \ge \frac{\ln (4-x)}{\ln 5}$$ 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 $$[3,4)$$.