Solve logarithmic equation $\log_{\frac{x}{5}}(x^2-8x+16)\geq 0$

Question

Find $x$ from logarithmic equation: $$\log_{\frac{x}{5}}(x^2-8x+16)\geq 0 $$ This is how I tried: $$x^2-8x+16>0$$ $$ (x-4)^2>0 \implies x \not = 4$$ then $$\log_{\frac{x}{5}}(x^2-8x+16)\geq \log_{\frac{x}{5}}(\frac{x}{5})^0 $$ because of base $\frac{x}{5}$, we assume $x \not\in (-5,5)$, then $$x^2-8x+16 \geq 1$$ $$ (x-3)(x-5) \geq 0 \implies$$ $$ \implies x \in {(- \infty,-5) \cup (5, \infty)} \cap x\not = 4 $$ But this is wrong, because the right solution is $$x \in {(3,4) \cup (4,6)} $$

I'm sorry if I used the wrong terms, English is not my native language.

Answer 1

Given $$\displaystyle \log_{\frac{x}{5}}(x^2-8x+16)\geq 0\;,$$ Here function is defined when $\displaystyle \frac{x}{5}>0$ and $\displaystyle \frac{x}{5}\neq 1$

and $(x-4)^2>0$. So we get $x>0$ and $x\neq 5$ and $x\neq 4$

If $$\displaystyle \; \bullet\; \frac{x}{5}>1\Rightarrow x>5\;,$$ Then $$\displaystyle \log_{\frac{x}{5}}(x^2-8x+16)\geq 0\Rightarrow (x^2-8x+16)\geq 1$$

So we get $$\displaystyle x^2-8x+15\geq 0\Rightarrow (x-3)(x-5)\geq 0$$

So we get $x>5$

If $$\displaystyle \; \bullet 0<\frac{x}{5}<1\Rightarrow 0<x<5\;,$$ Then $$\displaystyle \log_{\frac{x}{5}}(x^2-8x+16)\geq 0\Rightarrow (x^2-8x+16)\leq 1$$

So we get $$\displaystyle (x-3)(x-5)\leq 0$$

So $$3\leq x<5-\left\{4\right\}$$

So our final Solution is $$\displaystyle x\in \left[3,4\right)\cup \left(4,5\right)\cup \left(5,\infty\right)$$

Answer 2

$$\log_{\frac{x}{5}}(x^2-8x+16)\geq 0$$ $$=\frac{\log{(x^2-8x+16)}}{\log\frac{x}{5}}\geq0$$ So $x$ is non-positive. And now raising both sides on the base $10$. $$x^2-8x+16\geq1$$ $$x^2-8x+15\geq0$$ $$(x-5)(x-3)\geq0$$ $$x\geq5; \ x\geq3$$ But $x$ cannot be $5$. So answer is $[3,5)\cup(5,\infty)$