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

Dec 24, 2015

The step by step explanation is given below.

#### Explanation:

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

Let us try to get this as $\log \left(A\right) = \log \left(B\right)$ so that we can remove the logs and equate $A = B$

$2 = 2 \log \left(10\right) = \log \left({10}^{2}\right) = \log \left(100\right)$

Our equation would become
$\log \left({x}^{2} + 4\right) - \log \left(x + 2\right) = \log \left(100\right) + \log \left(x - 2\right)$
$\log \left(\frac{{x}^{2} + 4}{x + 2}\right) = \log \left(100 \left(x - 2\right)\right)$

Rule: $\log \left(A\right) - \log \left(B\right) = \log \left(\frac{A}{B}\right)$ and $\log \left(A\right) + \log \left(B\right) = \log \left(A B\right)$

$\frac{{x}^{2} + 4}{x + 2} = 100 \left(x - 2\right)$
$\left({x}^{2} + 4\right) = 100 \left(x - 2\right) \left(x + 2\right)$ this is got by multiplying both sides by (x+2)
${x}^{2} + 4 = 100 \left({x}^{2} - 4\right)$
${x}^{2} + 4 = 100 {x}^{2} - 400$

Add $400$ to both the sides.
${x}^{2} + 404 = 100 {x}^{2}$
Subtract x^2 from both the sides.
$404 = 99 {x}^{2}$
Divide by 99 on both the sides to isolate x^2

${x}^{2} = \frac{404}{99}$

$x = \pm \sqrt{\frac{404}{99}}$

$x = \sqrt{\frac{404}{99}}$ is the solution

$x = - \sqrt{\frac{404}{99}}$ does not satisfy the original equation and is discarded.

$x = \sqrt{\frac{404}{99}}$ Answer

Note: if you want an approximate value, use calculator and round off to the required decimal places.