Let $f:\mathbb{R}\to\mathbb{R}$ be a continuous function which satisfies $f(x)=\int_{0}^{x}f(t)\,dt$. Then the value of $f(\ln 5)$ is

$f(x)=\int_0^x f(t)\,dt$

$f \text{ continuous on } \mathbb{R}$

$\text{To find: } f(\ln 5)$

$\dfrac{d}{dx}\left[\int_0^x f(t)\,dt\right]=f(x) \;\Rightarrow\; f'(x)=f(x)$

$f \text{ continuous} \;\Rightarrow\; \text{FTC applies}$

$x=0:\ f(0)=\int_0^0 f(t)\,dt=0 \;\Rightarrow\; f(0)=0$

$\text{Plan: solve } f'=f \text{ with } f(0)=0$

$\dfrac{df}{dx}=f \;\Rightarrow\; \dfrac{df}{f}=dx \;\Rightarrow\; \ln|f|=x+C$

$f(x)=Ce^{x}$

$f(0)=Ce^{0}=C=0 \;\Rightarrow\; C=0$

$f(x)=0 \quad \text{for all } x$

$f\equiv 0:\ \int_0^x 0\,dt=0=f(x)\ \checkmark$

$f(\ln 5)=0$