$\text { Let } \frac{x}{y}=t \Rightarrow x=t y$
$\frac{d x}{d y}=t+y \frac{d t}{d y}$
$t+y \frac{d t}{d y}=t(\ln (t)+1)$
$y \frac{d t}{d y}=t \ln (t) \Rightarrow \frac{d t}{t \ln (t)}=\frac{d y}{y}$
$\Rightarrow \int \frac{d t}{t \cdot \ln (t)}=\int \frac{d y}{y}$
$\Rightarrow \int \frac{d p}{p}=\int \frac{d y}{y} \quad$ $\text { let } \ln t=p$
$\Rightarrow \ln p=\ln y+c$
$\ln (\ln t)=\ln y+c$
$\ln \left(\ln \left(\frac{x}{y}\right)\right)=\ln y+c$
$\text { at } x=e, y=1$
$\ln \left(\ln \left(\frac{e}{1}\right)\right)=\ln (1)+c \Rightarrow c=0$
$\ln \left|\ln \left(\frac{x}{y}\right)\right|=\ln y$
$\left|\ln \left(\frac{x}{y}\right)\right|=e^{\ln y}$
$\left|\ln \left(\frac{x}{y}\right)\right|=y$
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$(A)$ $\frac{2 \log _3 2}{2 \log _3 2-1}$ $(B)$ $\frac{2}{2-\log _2 3}$ $(C)$ $\frac{1}{1-\log _4 3}$ $(D)$ $\frac{2 \log _2 3}{2 \log _2 3-1}$