1. The problem is to find the value of $x$ that satisfies the equation $\ln x + \log x = 5$. 2. Here, $\ln x$ is the natural logarithm (log base $e$) and $\log x$ is the common logarithm (log base 10). 3. We use the change of base formula: $\log x = \frac{\ln x}{\ln 10}$. 4. Substitute into the equation: $$\ln x + \frac{\ln x}{\ln 10} = 5$$ 5. Factor out $\ln x$: $$\ln x \left(1 + \frac{1}{\ln 10}\right) = 5$$ 6. Simplify the factor: $$1 + \frac{1}{\ln 10} = \frac{\ln 10 + 1}{\ln 10}$$ 7. So, $$\ln x \cdot \frac{\ln 10 + 1}{\ln 10} = 5$$ 8. Solve for $\ln x$: $$\ln x = 5 \cdot \frac{\ln 10}{\ln 10 + 1}$$ 9. Calculate numerically using $\ln 10 \approx 2.302585$: $$\ln x = 5 \cdot \frac{2.302585}{2.302585 + 1} = 5 \cdot \frac{2.302585}{3.302585} \approx 5 \cdot 0.697 = 3.485$$ 10. Exponentiate both sides to solve for $x$: $$x = e^{3.485}$$ 11. Calculate $x$: $$x \approx 32.656$$ 12. Therefore, the solution is $x \approx 32.656$, which corresponds to option B.