How do you solve #ln (x – 2) + ln (x + 2) = ln 5#?

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Answer 1 · 2015-09-09T18:11:17

Use properties of #ln# to find #ln(x^2-4) = ln(5)# and hence find solution: #x = 3#

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

#= ln(5)#

So #x^2-4 = 5#

Add #4# to both sides to get: #x^2 = 9#

So #x = +-3#

If #x = -3#, then #x - 2 < 0# and #x + 2 < 0#, so neither #ln(x-2)# nor #ln(x+2)# have Real values.

If #x = 3# then:

#ln(x-2) + ln(x+2) = ln(1) + ln(5) = 0 + ln(5) = ln(5)#

So the only solution is #x = 3#

Answer 2 · 2015-09-09T18:12:50

Well you can write it as follows

#ln(x-2)+ln(x+2)=ln5=>ln(x^2-4)=ln5=>x^2-4=5=>x^2=9=>x=-3,x=+3#

But x-2>0 so the solution is x=3