# How do you solve ln x - 4 ln 3 = ln (5 / x)?

Aug 13, 2015

$\textcolor{red}{x = 9 \sqrt{5}}$

#### Explanation:

$\ln x - 4 \ln 3 = \ln \left(\frac{5}{x}\right)$

Recall that $a \ln b = \ln \left({a}^{b}\right)$, so

$\ln x - \ln {3}^{4} = \ln \left(\frac{5}{x}\right)$

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

$\ln \left(\frac{x}{3} ^ 4\right) = \ln \left(\frac{5}{x}\right)$

Convert the logarithmic equation to an exponential equation.

${e}^{\ln} \left(\frac{x}{3} ^ 4\right) = {e}^{\ln} \left(\frac{5}{x}\right)$

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

$\frac{x}{3} ^ 4 = \frac{5}{x}$

x^2=5×3^4

x=±sqrt(5×3^4) =±sqrt5×sqrt(3^4)=±sqrt5×3^2

$x = 9 \sqrt{5}$ and $x = - 9 \sqrt{5}$ are possible solutions.

Check:

$\ln x - 4 \ln 3 = \ln \left(\frac{5}{x}\right)$

If $x = 9 \sqrt{5}$,

$\ln \left(9 \sqrt{5}\right) - 4 \ln 3 = \ln \left(\frac{5}{9 \sqrt{5}}\right)$

$\ln 9 + \ln \sqrt{5} - \ln {3}^{4} = \ln 5 - \ln \left(9 \sqrt{5}\right)$

$\ln 9 + \ln \sqrt{5} - \ln {\left({3}^{2}\right)}^{2} = \ln {\left(\sqrt{5}\right)}^{2} - \ln 9 - \ln \sqrt{5}$

$\ln 9 + \ln \sqrt{5} - 2 \ln 9 = 2 \ln \sqrt{5} - \ln 9 - \ln \sqrt{5}$

$\ln \sqrt{5} - \ln 9 = \ln \sqrt{5} - \ln 9$

$x = 9 \sqrt{5}$ is a solution.

If $x = - 9 \sqrt{5}$,

$\ln \left(- 9 \sqrt{5}\right) - 4 \ln 3 = \ln \left(\frac{5}{- 9 \sqrt{5}}\right) = \ln 5 - \ln \left(- 9 \sqrt{5}\right)$

But $\ln \left(- 9 \sqrt{5}\right)$ is not defined.

$x = - 9 \sqrt{5}$ is not a solution.