[tex]f(x)=e^x[/tex] is continuous and differentiable on its entire domain, so the mean value theorem holds for any interval you pick. We have
[tex]f'(x)=e^x[/tex]
and the MVT says there is some [tex]c\in(0,\ln15)[/tex] such that
[tex]f'(c)=\dfrac{f(\ln15)-f(0)}{\ln15-0}[/tex]
So we have
[tex]e^c=\dfrac{15-1}{\ln15}=\dfrac{14}{\ln15}[/tex]
[tex]\implies c=\ln\dfrac{14}{\ln15}\approx1.642[/tex]