Answer
a. 5
b. e
Work Step by Step
$\displaystyle \int_{1}^{x}\frac{1}{t}dt=$ Log Rule = $\ln|t|_{1}^{x}$
$= \ln|x|-\ln 1$
the integral bounds suggest $x > 1 > 0$, so we write
$=\ln x$
a.
$\ln x=\ln 5\qquad $/... apply $e^{(..)}$ to both sides
$x=5$
b.
$\ln x=1\qquad $/... apply: $\ln e =1$
$\ln x=\ln e\qquad $/... apply $e^{(..)}$ to both sides
$x=e$
