# How do you solve ln(x)=3+ln(x-3)?

Aug 13, 2015

$\textcolor{red}{x = \frac{3 {e}^{3}}{{e}^{3} - 1}}$

#### Explanation:

$\ln x = 3 + \ln \left(x - 3\right)$

Subtract $\ln \left(x - 3\right)$ from each side.

$\ln x - \ln \left(x - 3\right) = 3$

Recall that $\ln a - \ln b = \ln \left(\frac{a}{b}\right)$.

$\ln \left(\frac{x}{x - 3}\right) = 3$

Convert the logarithmic equation to an exponential equation.

${e}^{\ln} \left(\frac{x}{x - 3}\right) = {e}^{3}$

Remember that ${e}^{\ln} x = x$, so

$\frac{x}{x - 3} = {e}^{3}$

$x = {e}^{3} \left(x - 3\right) = x {e}^{3} - 3 {e}^{3}$

$x {e}^{3} - x = 3 {e}^{3}$

$x \left({e}^{3} - 1\right) = 3 {e}^{3}$

$x = \frac{3 {e}^{3}}{{e}^{3} - 1}$

Check:

$\ln x = 3 + \ln \left(x - 3\right)$

If $x = \frac{3 {e}^{3}}{{e}^{3} - 1}$,

$\ln \left(\frac{3 {e}^{3}}{{e}^{3} - 1}\right) = 3 + \ln \left(\frac{3 {e}^{3}}{{e}^{3} - 1} - 3\right)$

$\ln \left(3 {e}^{3}\right) - \ln \left({e}^{3} - 1\right) = 3 + \ln \left(\frac{3 {e}^{3} - 3 \left({e}^{3} - 1\right)}{{e}^{3} - 1}\right)$

$\ln \left(3 {e}^{3}\right) - \ln \left({e}^{3} - 1\right) = 3 + \ln \left(\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{3 {e}^{3}}}} - \textcolor{red}{\cancel{\textcolor{b l a c k}{3 {e}^{3}}}} + 3}{{e}^{3} - 1}\right)$

$\ln \left(3 {e}^{3}\right) - \textcolor{red}{\cancel{\textcolor{b l a c k}{\ln \left({e}^{3} - 1\right)}}} = 3 + \ln 3 - \textcolor{red}{\cancel{\textcolor{b l a c k}{\ln \left({e}^{3} - 1\right)}}}$

$\ln 3 + 3 = 3 + \ln 3$

$x = \frac{3 {e}^{3}}{{e}^{3} - 1}$ is a solution.