# How do you solve ln(2x^2-2)-ln9=ln80?

Aug 18, 2016

The Soln.$: x = \pm 19$.

#### Explanation:

$\ln \left(2 {x}^{2} - 2\right) - \ln 9 = \ln 80$

$\therefore \ln \left(2 {x}^{2} - 2\right) = \ln 9 + \ln 80 = \ln \left(9 \cdot 80\right)$

Since, $\ln$ fun. is $1 - 1$, we have,

$2 {x}^{2} - 2 = 9 \cdot 80$

$\Rightarrow 2 \left({x}^{2} - 1\right) = 9 \cdot 80$

$\Rightarrow {x}^{2} - 1 = 9 \cdot 40 = 360$

$\Rightarrow {x}^{2} = 361$

$\Rightarrow x = \pm \sqrt{361} = \pm 19$

$x = + 19 , \mathmr{and} , x = - 19$ satisfy the given eqn.

Hence, the Soln. $x = \pm 19$.

Aug 18, 2016

$x = \pm 19$

#### Explanation:

If the logs are being subtracted, the numbers are being divided.

We can condense two ln terms into one.

$\ln \left(2 {x}^{2} - 2\right) - \ln 9 = \ln 80$

$\ln \left(\frac{2 {x}^{2} - 2}{9}\right) = \ln 80$

$\therefore \frac{2 {x}^{2} - 2}{9} = 80 \text{ if } \ln \left(\frac{a}{b}\right) = \ln c \Rightarrow \frac{a}{b} = c$

$2 {x}^{2} - 2 = 720$

$2 {x}^{2} = 722$

${x}^{2} = 361$

$x = \pm \sqrt{361}$

$x = \pm 19$