Question

2/(8^x-10)>=4/(8^x-8)

Answer 1

$\frac{2}{8^{x}-10} \geq \frac{4}{8^{x}-8}\\\\t=8^{x}\ \textgreater \ 0\; \; \frac{1}{t-10}-\frac{2}{t-8} \geq 0\\\\\frac{t-8-2t+20}{(t-10)(t-8)} \geq 0\\\\\frac{-(t-12)}{(t-10)(t-8)} \geq 0\\\\\frac{t-12}{(t-10)(t-8)} \leq 0\\\\---(8)+++(10)---[12]+++\\\\t\in (-\infty,8)U(10,12\, ]\\\\t\ \textgreater \ 0\; \; \to \; t\in (0,8)U(10,12\, ]$

$\left \{ {{0 \ \textless \ 8^{x}\ \textless \ 8} \atop {10\ \textless \ 8^{x} \leq 12}} \right. \; \left \{ {{x\ \textless \ 1} \atop {log_810\ \textless \ x \leq log_812}} \right. \\\\x\in(-\infty,1)U(log_810,log_812\, ]$