Solve the differential equation \[ \frac{d x}{d t}=\frac{x \ln x}{t} \] Assume \( x, t>0 \), and use the initial condition \( x(1)=5 \) \( x= \)

We start with the differential equation \[ \frac{dx}{dt} = \frac{x \ln x}{t}. \] Since \( x>0 \) and \( t>0 \), we can separate variables by rewriting the equation as \[ \frac{dx}{x \ln x} = \frac{dt}{t}. \] 1. **Integrate both sides:** \[ \int \frac{dx}{x \ln x} = \int \frac{dt}{t}. \] 2. **Solve the left integral by substitution:** Let \[ u = \ln x \quad \text{so that} \quad du = \frac{dx}{x}. \] Then, the integral becomes \[ \int \frac{du}{u} = \ln |u| + C_1 = \ln |\ln x| + C_1. \] 3. **Solve the right integral:** \[ \int \frac{dt}{t} = \ln |t| + C_2. \] 4. **Combine the results:** Equate the two integrals (we can combine constants into a single constant \( C \)): \[ \ln |\ln x| = \ln |t| + C. \] Since \( x > 0 \) and given \( x(1)=5>1 \), we have \( \ln x >0 \), so we can drop the absolute values: \[ \ln (\ln x) = \ln t + C. \] 5. **Exponentiate both sides to solve for \(\ln x\):** \[ \ln x = e^{\ln t + C} = e^C \cdot t. \] Denote \( K = e^C \) (where \( K > 0 \)), then \[ \ln x = Kt. \] 6. **Determine the constant \( K \) using the initial condition \( x(1)=5 \):** Substitute \( t = 1 \) and \( x = 5 \) into the equation: \[ \ln 5 = K \cdot 1 \quad \Longrightarrow \quad K = \ln 5. \] Therefore, the equation becomes \[ \ln x = (\ln 5)t. \] 7. **Solve for \( x \):** Exponentiate both sides: \[ x = e^{(\ln 5)t} = \left(e^{\ln 5}\right)^t = 5^t. \] Thus, the solution to the differential equation is \[ x(t)=5^t. \]