Question

Solve $\frac{1-5^x}{7^{-x}-7}\ge 0$,
what is the maximum negative value x can take?

Answer 1

$g\left(x\right)=\frac{1-5^x}{7^{-x}-7}\ge 0$. Now $5^x-1=0$
$\Rightarrow$ $x=0$ and $7^{-x}-7=0$ $\Rightarrow$ x=-1
$xϵ(-\infty ,-1)\cup [0,\infty )$
Ans: -1