1. The problem is to factor out $\ln(x)$ from an expression where it appears as a common factor. 2. Factoring means rewriting an expression as a product of factors. If $\ln(x)$ appears in every term, you can factor it out using the distributive property: $a b + a c = a(b + c)$. 3. For example, if you have $\ln(x) \cdot f(x) + \ln(x) \cdot g(x)$, you can factor out $\ln(x)$ as: $$\ln(x) \cdot f(x) + \ln(x) \cdot g(x) = \ln(x) (f(x) + g(x))$$ 4. This works because $\ln(x)$ is common to both terms, so you group it outside the parentheses. 5. Always check that $\ln(x)$ is indeed a factor of every term before factoring it out. 6. If you have a specific expression, please provide it for exact factoring steps.