How do you solve lnx=3+ln(x-5)?

Mar 9, 2016

$x = \frac{5}{1 - {e}^{-} 3}$

Explanation:

Transposing term in $\ln x = 3 + \ln \left(x - 5\right)$

$\ln x - \ln \left(x - 5\right) = 3$ or

$\ln \left(\frac{x}{x - 5}\right) = 3$ - as $\ln a - \ln b = \ln \left(\frac{a}{b}\right)$

Hence, as $\ln a = b \Rightarrow a = {e}^{b}$

$\left(\frac{x}{x - 5}\right) = {e}^{3}$ or

$x = \left(x - 5\right) {e}^{3}$ or

$\left({e}^{3} - 1\right) x = 5 {e}^{3}$ or

$x = \frac{5 {e}^{3}}{{e}^{3} - 1}$ or $x = \frac{5}{1 - {e}^{-} 3}$