# How do you solve ln(5-2x^2)+ln9=ln43?

May 27, 2018

$x = \pm \frac{1}{3}$

#### Explanation:

We have
$\ln \left(5 - 2 {x}^{2}\right) = \ln \left(43\right) - \ln \left(9\right)$

$\ln \left(5 - 2 {x}^{2}\right) = \ln \left(\frac{43}{9}\right)$

so

$5 - 2 {x}^{2} = \frac{43}{9}$

$5 - \frac{43}{9} = 2 {x}^{2}$

$\frac{2}{9} = 2 {x}^{2}$

$\frac{1}{9} = {x}^{2}$

May 27, 2018

See explanation below

#### Explanation:

The goal is to get a expresion $\log A = \log B$. By inyectivity of logarithm we arrive to $A = B$

In our case: $\ln \left(5 - 2 {x}^{2}\right) + \ln 9 = \ln 43$

$\ln \left(5 - 2 {x}^{2}\right) = \ln 43 - \ln 9 = \ln \frac{43}{9}$

Then $5 - 2 {x}^{2} = \frac{43}{9}$ or equivalent

$5 - \frac{43}{9} = 2 {x}^{2}$

$\frac{2}{9} = 2 {x}^{2}$

$\frac{1}{9} = {x}^{2}$

$x = \pm \frac{1}{3}$

It is obvious that both solution are valid

May 27, 2018

$x = \pm \frac{1}{3}$

#### Explanation:

Given: $\ln \left(5 - 2 {x}^{2}\right) + \ln 9 = \ln 43$

Use the logarithm property $\ln a + \ln b = \ln \left(a \cdot b\right)$

$\ln \left(\left(5 - 2 {x}^{2}\right) 9\right) = \ln 43$

$\ln \left(45 - 18 {x}^{2}\right) = \ln 43$

Exponentiate both sides and use the logarithm property ${e}^{\ln} x = x$

${e}^{\ln \left(45 - 18 {x}^{2}\right)} = {e}^{\ln 43}$

$45 - 18 {x}^{2} = 43$

$45 - 43 = 18 {x}^{2}$

$2 = 18 {x}^{2}$

$\frac{2}{18} = \frac{1}{9} = {x}^{2}$

$x = \pm \frac{1}{3}$

CHECK to see if the answers work in the problem (must be ln of a positive number)#:

$\ln \left(5 - 2 {\left(\frac{1}{3}\right)}^{2}\right) = \ln \left(5 - \frac{2}{9}\right) > 0$

$\ln \left(5 - 2 {\left(- \frac{1}{3}\right)}^{2}\right) = \ln \left(5 - \frac{2}{9}\right) > 0$