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

2 Answers
Jul 28, 2016

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)

Jul 28, 2016

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.