# How do you solve ln(x-2)-ln(x+2)=2?

##### 2 Answers
Jul 28, 2016

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

#### Explanation:

This problem requires some clever manipulation and knowledge of logarithm/exponential identities but is easily solved with the right toolset.

We start by taking the exponential function of both sides. That is, replacing each side with $e$ to the power of that side:

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

Now we use the fact that subtraction in an exponential is the same as division outside it:

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

Now we use the fact that the exponent of the natual logarithm of any number is just that number:

${e}^{\ln \left(x - 2\right)} / {e}^{\ln \left(x + 2\right)} = \frac{x - 2}{x + 2} = {e}^{2}$

From here, the solution proceeds by standard algebra:

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

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

$x - 2 = {e}^{2} x + 2 {e}^{2}$

$x - {e}^{2} x = 2 {e}^{2} + 2$

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

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

Jul 28, 2016

No solution, if we are only considering Real values.

Upon examining Complex solutions, we find that $x = \frac{2 \left({e}^{2} + 1\right)}{1 - {e}^{2}}$.

#### Explanation:

From the outset, we can apply the logarithm rule:

$\textcolor{red}{\overline{\underline{|}} \textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\log \left(A\right) - \log \left(B\right) = \log \left(\frac{A}{B}\right)} \textcolor{w h i t e}{\frac{a}{a}} |}$

Yielding the simpler equation:

$\ln \left(\frac{x - 2}{x + 2}\right) = 2$

From here, use the relation:

From the outset, we can apply the logarithm rule:

$\textcolor{b l u e}{\overline{\underline{|}} \textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\ln \left(A\right) = B \text{ "hArr" } A = {e}^{B}} \textcolor{w h i t e}{\frac{a}{a}} |}$

This can also be viewed as exponentiating both sides with a base of $e$.

Thus:

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

Using the algebra already outlined in the other answer:

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

Distributing:

$x - 2 = {e}^{2} x + 2 {e}^{2}$

Grouping the $x$ terms:

$x - {e}^{2} x = 2 {e}^{2} + 2$

Factoring:

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

Dividing:

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

IMPORTANT NOTE!

The value of $\frac{2 \left({e}^{2} + 1\right)}{1 - {e}^{2}}$ is approximately $- 2.6261$. Thus, in $\ln \left(x - 2\right)$ and $\ln \left(x + 2\right)$, the argument of the natural logarithm is negative. For the typical Real-valued logarithm function, this is prohibited.

Thus the value of $x = \frac{2 \left({e}^{2} + 1\right)}{1 - {e}^{2}}$ is only allowed if we are assuming to be working with a Complex-valued logarithm.