Answer
a) $\frac{1}{3}\lt x\lt\frac{1+\ln2}{3}$
b) $x\gt\frac{1}{e}$
Work Step by Step
a) $1\lt e^{3x-1}\lt2$
$0=\ln1\lt3x-1\lt\ln2$
$1\lt3x\lt1+\ln2$
$\frac{1}{3}\lt x\lt\frac{1+\ln2}{3}$
b)$1-2\ln x\lt3$
$2\ln x\gt-2$
$\ln x\gt-1$
$x\gt e^{-1}=\frac{1}{e}$
