Answer
$x=2$
Work Step by Step
LHS = $3\displaystyle \int_{1}^{x}\frac{1}{t}dt=$ Log Rule = $3\ln|t|_{1}^{x}$
$LHS= 3\ln x-3\ln 1=3\ln x$
$RHS$ = log rule = $\ln|t|_{1/4}^{x}$ = $\displaystyle \ln x-\ln\frac{1}{4}$
$=\ln x-\ln 4^{-1}=\ln x+\ln 4$
$LHS=RHS$ is now a logarithmic equation...
$3\ln x=\ln x+\ln 4\qquad/-\ln x$
$ 2\ln x=\ln 2^{2}\qquad$ /RHS ... log of a power...
$ 2\ln x=2\ln 2\qquad$ .../$\div 2$
$\ln x=\ln 2\qquad $/... apply $e^{(..)}$ to both sides
$x=2$
