Question

Решите неравенство 3^x-2-(1/3)^-x+8=0

Answer 1

$3^{x-2}-(\frac{1}{3})^{-x}+8=0$
$3^{x-2}-(3^{-1})^{-x}=0-8=-8$
$3^{x-2}-3^x=-8$
$\frac{3^x}{3^2}-\frac{3^x\cdot3^2}{3^2}=-8$
$\frac{3^x-3^x\cdot3^2}{3^2}=-8$
$3^x-3^x\cdot9=-8\cdot3^2=-72<span>$
$-8\cdot3^x=-72$
$3^x=\frac{-72}{-8}=9$
$x=\log_39=2$