Solving an equation with two natural logarithms

Use the power property to bring the exponent out: $\ln (x^3)=3\ln x$. Then $$3\ln x-\ln x=4\Rightarrow 2\ln x=4\Rightarrow \ln x=2\Rightarrow x=e^2$$

Now the domain. The original equation contains $\ln x$, which is defined only for $x>0$. So a negative value can never be a solution, whatever the algebra suggests.

If you combine with the quotient property instead, you get $\ln \left(\frac{x^3}{x}\right)=\ln (x^2)=4$, so $x^2=e^4$, which looks like it gives $x=\pm e^2$. But $-e^2$ leaves $\ln x$ without a value in the original equation, so it is rejected.

Only $x=e^2$ works, choice C. Check: $\ln (e^6)-\ln (e^2)=6-2=4$.

Each choice

• A. This solves $x^2=4$, as if the logarithms could simply be dropped. Undoing $\ln$ needs exponentiation with base $e$, giving $x^2=e^4$. The value $-2$ would also fail the domain.
• B. This keeps both roots of $x^2=e^4$ and skips the domain check. $\ln x$ has no value at $-e^2$, so that value cannot satisfy the original equation.
• C (correct). The equation simplifies to $\ln x=2$, so $x=e^2$, and the domain of $\ln x$ rules out the negative root.
• D. Setting $\ln x=4$ drops the factor of 2 from $3\ln x-\ln x$. Checking it: $\ln (e^{12})-\ln (e^4)=12-4=8$, not 4.