How do you solve #ln(x) + ln(x-2) = ln(x+4)#?

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Answer 1 · 2015-10-21T11:32:56

Use properties of logs to derive a quadratic equation in #x#, one of whose roots is a valid solution to the original equation.

#x = 4#

#ln(x+4) = ln(x) + ln(x-2) = ln(x(x-2))#

So

#x+4 = x(x-2) = x^2 - 2x#

Subtract #x+4# from both ends to get:

#0 = x^2-3x-4 = (x-4)(x+1)#

So #x = -1# or #x = 4#

Discard #x = -1# since both #ln(-1)# and #ln(-1-2)# are not defined.

So the only solution is #x = 4#