Finding the value of $x$ if $\ln(x)+\ln(e)=\ln(18)$

Question

$\ln(x)+\ln(e)=\ln(18)$

The first step I did was making $\ln(e)=1 $ and then bring $\ln(x) $ to the other side of the equation.

$1=\ln(18) -\ln(x) $

$1=\ln(18 - x) $

Then I raise the numbers to be exponents of $e$, but I believe perhaps there is a different way, or I'm doing the incorrect way, because the problem should be used without a calculator, according to my teacher. But here are the following steps I did:

$e^1=(18-x) $

$x=18+e $

Is this the correct answer?

Answer 1

No, $18+e$ is clearly wrong (perhaps, there could theoretically be more than one correct answer, but in fact, there is only one correct answer, as I will show now). If $\operatorname{ln}(x)+\operatorname{ln}(e)=\operatorname{ln}(18)$, then applying $e^x$ to both sides gives $xe=18$, so $x=\frac{18}{e} \neq 18+e$. There is no other correct solution for $x$.

Answer 2

$\ln(x)+\ln(e)=\ln(18)$, so $\ln(x)=\ln(18)-\ln(e) = \ln(\frac{18}{e})$ and it follows that $x=\frac{18}{e}$.

Alternatively: $\ln(x)+\ln(e)=\ln(ex) = \ln(18)$ so $ex=18$ and we get the same solution.