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

##### 2 Answers

#### 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

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

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

From here, the solution proceeds by standard algebra:

No solution, if we are only considering Real values.

Upon examining Complex solutions, we find that

#### Explanation:

From the outset, we can apply the logarithm rule:

#color(red)(barul|color(white)(a/a)color(black)(log(A)-log(B)=log(A/B))color(white)(a/a)|)#

Yielding the simpler equation:

#ln((x-2)/(x+2))=2#

From here, use the relation:

From the outset, we can apply the logarithm rule:

#color(blue)(barul|color(white)(a/a)color(black)(ln(A)=B" "hArr" "A=e^B)color(white)(a/a)|)#

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

Thus:

#(x-2)/(x+2)=e^2#

Using the algebra already outlined in the other answer:

#x-2=e^2(x+2)#

Distributing:

#x-2=e^2x+2e^2#

Grouping the

#x-e^2x=2e^2+2#

Factoring:

#x(1-e^2)=2e^2+2#

Dividing:

#x=(2(e^2+1))/(1-e^2)#

**IMPORTANT NOTE!**

The value of

Thus the value of