# How do you solve #ln (x + 2) + ln (x - 2) = 0#?

##### 1 Answer

Apr 26, 2016

Take exponents of both sides to get a quadratic, the positive root of which is the solution:

#x = sqrt(5)#

#### Explanation:

Given:

#ln(x+2)+ln(x-2) = 0#

If we are dealing with Real logarithms then taking exponents of both sides we find:

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

That is:

#x^2-4 = 1#

So:

#x^2 = 5#

Hence:

#x = +-sqrt(5)#

We can discard the negative root since we require

So the remaining solution is

**Footnote**

If

#ln t = ln abs(t) + pi i#

[[ More generally

So we find:

#ln(-sqrt(5)+2) + ln(-sqrt(5)-2)#

#= ln(sqrt(5)-2) + pi i + ln(sqrt(5)+2) + pi i = 2 pi i#