I need help solving and understanding this problem too!!!

ANSWERD.EXPLANATIONIf [tex] ln(x) = y[/tex]Then solving for x will gives us:[tex] {e}^{ ln(x) } = {e}^{y} [/tex][tex]x = {e}^{y} [/tex]This first pair are equivalent:If[tex]log_{b}(N)= P[/tex]weThen,[tex] {b}^{log_{b}(N) = {b}^{ P}} [/tex][tex]N= {b}^{ P}[/tex]This pairs are also equivalent.If [tex]x = \sqrt{y} [/tex]Then we rewrite as an index to obtain;[tex]x = {y}^{ \frac{1}{2} } [/tex]This pair is also equivalent.[tex]log_{p}(N)= b[/tex]Then[tex] {p}^{log_{p}(N)} = {p}^{b} [/tex][tex] {p}^{b} = N[/tex]But this was not the pair given in the last option.The correct answer is D