Answer
$a)$ $\frac{1}{3}\lt x \lt \frac{\ln(2)+1}{3}$
$b)$ $x \gt e^{-1}$
Work Step by Step
$a$-
$$1 \lt e^{3x-1} \lt 2$$
$$\ln (1) \lt \ln(e^{3x-1}) \lt \ln(2)$$
$$0\lt 3x-1 \lt \ln(2)$$
$$1\lt 3x \lt \ln(2)+1$$
$$\frac{1}{3}\lt x \lt \frac{\ln(2)+1}{3}$$
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$b$-
$$1-2\ln x \lt 3$$
$$2\ln x-1 \gt -3$$
$$2\ln x \gt -3+1$$
$$2\ln x \gt -2$$
$$\ln x \gt -1$$
$$e^{\ln x} \gt e^{-1}$$
$$x \gt e^{-1}$$
