We are given that

\(X\sim N(\mu, \ \sigma^{2}) \ \text{and}\) \(Y-e^{X}\).

Observe that \(Y>0\) with porbability 1.

So that for ant \(y>0\) we have that

\({F}_{{Y}}{\left({y}\right)}={P}{\left({Y}\le{y}\right)}={P}{\left({e}^{x}\le{y}\right)}={P}{\left({X}\le \log{{\left({y}\right)}}\right)}={F}_{{X}}{\left( \log{{\left({y}\right)}}\right)}\)

where \(F_X\) is CDF of X . By differentiating, we have that

\({{f}_{{Y}}{\left({y}\right)}}=\frac{d}{{\left.{d}{y}\right.}}{F}_{{y}}{\left({y}\right)}={{f}_{{X}}{\left( \log{{\left({y}\right)}}\right)}}\). \(\frac{1}{{y}}=\frac{1}{{\sigma\sqrt{{2}}\pi}} \exp{{\left(-\frac{{{\left( \log{{u}}-\mu\right)}^{2}}}{{{2}\sigma^{2}}}\right)}}\cdot\frac{1}{{y}}\)