# How do you solve log(x^2+4)-log(x+2)=2+log(x-2)?

Aug 8, 2018

#### Explanation:

Here,

$\log \left({x}^{2} + 4\right) - \log \left(x + 2\right) = 2 + \log \left(x - 2\right)$

$\implies \log \left({x}^{2} + 4\right) - \log \left(x + 2\right) - \log \left(x - 2\right) = 2$

$\implies \log \left({x}^{2} + 4\right) - \left\{\log \left(x + 2\right) + \log \left(x - 2\right)\right\} = 2$

Using : $\log M + \log N = \log \left(M N\right)$

$\implies \log \left({x}^{2} + 4\right) - \log \left[\left(x + 2\right) \left(x - 2\right)\right] = 2$

$\implies \log \left({x}^{2} + 4\right) - \log \left({x}^{2} - 4\right) = 2$

Using : $\log M - \log N = \log \left(\frac{M}{N}\right)$

$\log \left(\frac{{x}^{2} + 4}{{x}^{2} - 4}\right) = 2$

$\left(i\right)$If it is common logarithm-logarithm to base $10$ ,then

${\log}_{10} \left(\frac{{x}^{2} + 4}{{x}^{2} - 4}\right) = 2$

:.(x^2+4)/(x^2-4)=10^2,where,x^2!=4=>color(red)(x!=+-2

${x}^{2} + 4 = 100 \left({x}^{2} - 4\right)$

$\therefore 100 {x}^{2} - 400 - {x}^{2} - 4 = 0$

$99 {x}^{2} - 404 = 0$

$\therefore {x}^{2} = \frac{404}{99} \approx 4.08$

$\therefore x = \pm \sqrt{4.08} \approx 2.02$

But, $x \approx - 2.02$ will make $\log \left(x - 2\right)$ meaningless.

$\therefore x \approx 2.02$

Note that most of the textbooks use $\log x$ as,
logarithm to base 10

$\left(i i\right)$If it is color(blue)"natural logarithm ??-logarithm to base e " ,then

${\log}_{e} \left(\frac{{x}^{2} + 4}{{x}^{2} - 4}\right) = 2$

:.(x^2+4)/(x^2-4)=e^2,where,x^2!=4=>color(red)(x!=+-2

${x}^{2} + 4 = {e}^{2} \left({x}^{2} - 4\right)$

$\therefore {e}^{2} {x}^{2} - 4 {e}^{2} - {x}^{2} - 4 = 0$

$\therefore {e}^{2} {x}^{2} - {x}^{2} = 4 {e}^{2} + 4$

$\therefore {x}^{2} \left({e}^{2} - 1\right) = 4 \left({e}^{2} + 1\right)$

$\therefore {x}^{2} = 4 \left(\frac{{e}^{2} + 1}{{e}^{2} - 1}\right)$

$\therefore x = 2 \sqrt{\frac{{e}^{2} + 1}{{e}^{2} - 1}}$