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 x+2 > 0 and x - 2 > 0.

So the remaining solution is x = sqrt(5)

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Footnote

x=-sqrt(5) is not a solution even if we consider Complex logarithms.

If t < 0 then the principal Complex natural logarithm has the value:

ln t = ln abs(t) + pi i

[[ More generally ln z = ln abs(z) + Arg(z) i ]]

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