Let f(x) be a real differentiable function such that f(0)=1 and f(x+y)=f(x) f^prime(y)+f^prime(x) f(y) for all x, y ∈ R. Then Σ_ n=1^100 log _ e f(…

$\because f(x+y)=f(x) \cdot f(y)+f(x) \cdot f(y), \forall x, y \in R$∵f(x+y)=f(x)⋅f(y)+f(x)⋅f(y),∀x,y∈R ....(i) And $f(0)=1$f(0)=1 ....(ii) Now replace $x$x by zero and $y$y by zero we get $\begin{aligned}
& f(0)=f(0) f(0)+f(0) f(0) \
& 1=f(0)+f(0) \
& \therefore \quad f^{\prime}(0)=\frac{1}{2} ...(iii)
\end{aligned}$Now replace$Nowreplacey$by zero in equation (i), we get$byzeroinequation(i),wegetf(x)=\frac{1}{2} f(x)+f^{\prime}(x)$or,$or,\frac{1}{2} f(x)=f^{\prime}(x)$then$then\frac{f^{\prime}(x)}{f(x)}=\frac{1}{2}$hence$hence\ln |f(x)|=\frac{x}{2}+c$Put$Putx=0$, we get$,wegetc=0$$\therefore \quad \ln |f(x)|=\frac{x}{2}$Then$Then\sum_{n=1}^{100} \ln (f(\eta))=\left(\frac{1}{2}+\frac{2}{2}+\frac{3}{2}+\ldots+\frac{100}{2}\right)$$=\frac{5050}{2}=2525$