Answer
$\frac{1}{3} \lt x \lt \frac{\ln (2)+1}{3}$
Work Step by Step
$1 \lt e^{3x-1} \lt 2$
$\ln 1 \lt \ln (e^{3x-1}) \lt \ln (2)$
$\ln 1 \lt 3x-1 \lt \ln (2)$
$0 \lt 3x-1 \lt \ln (2)$
$0+1 \lt 3x-1+1 \lt \ln (2)+1$
$0+1 \lt 3x \lt \ln (2)+1$
$1 \lt 3x \lt \ln (2)+1$
$\frac{1}{3} \lt \frac{3x}{3} \lt \frac{\ln (2)+1}{3}$
$\frac{1}{3} \lt x \lt \frac{\ln (2)+1}{3}$
Therefore, the solution is:
$$\frac{1}{3} \lt x \lt \frac{\ln (2)+1}{3}$$
