Value of Continuous Function Defined by Definite Integral

To find the value of $f(\ln 5)$, we are given the equation:
$f(x) = \int_0^x f(t) \, dt$
where $f : \mathbb{R} \to \mathbb{R}$ is a continuous function.

1. Find the initial condition:
   Substituting $x = 0$ into the given equation:
   $f(0) = \int_0^0 f(t) \, dt = 0$

2. Differentiate both sides with respect to $x$:
   Since $f$ is continuous, by the Fundamental Theorem of Calculus, the integral is differentiable, which implies $f(x)$ is differentiable on $\mathbb{R}$. Differentiating both sides gives:
   $f'(x) = \frac{d}{dx} \left( \int_0^x f(t) \, dt \right) = f(x)$

3. Solve the differential equation:
   The equation is:
   $f'(x) - f(x) = 0$
   Multiplying both sides by the integrating factor $e^{-x}$:
   $\frac{d}{dx}\left( e^{-x} f(x) \right) = 0$
   Integrating both sides with respect to $x$:
   $e^{-x} f(x) = C$
   $f(x) = C e^x$
   where $C$ is a constant.

4. Apply the initial condition:
   Using $f(0) = 0$:
   $f(0) = C e^0 = C = 0$

   Thus, the only function satisfying the given condition is the zero function:
   $f(x) = 0 \quad \text{for all } x \in \mathbb{R}$

5. Calculate $f(\ln 5)$:
   $f(\ln 5) = 0$