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

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Answer 1 · 2016-04-26T16:56:13

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

#x = sqrt(5)#

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)#

#color(white)()#
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#