Question

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

Answer 1

`x = (2e^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(x-2) - ln(x+2)) = e^2`

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

`e^(ln(x-2) - ln(x+2)) = e^(ln(x-2))/e^(ln(x+2)) = e^2`

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

`e^(ln(x-2))/e^(ln(x+2)) = (x-2)/(x+2) = e^2`

From here, the solution proceeds by standard algebra:

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

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

`x - 2 = e^2 x + 2e^2`

`x - e^2 x = 2e^2 + 2`

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

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

Answer 2

No solution, if we are only considering Real values.

Upon examining Complex solutions, we find that `x=(2(e^2+1))/(1-e^2)`.

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 `e`.

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` terms:

`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 `(2(e^2+1))/(1-e^2)` is approximately `-2.6261`. Thus, in `ln(x-2)` and `ln(x+2)`, the argument of the natural logarithm is negative. For the typical Real-valued logarithm function, this is prohibited.

Thus the value of `x=(2(e^2+1))/(1-e^2)` is only allowed if we are assuming to be working with a Complex-valued logarithm.