Asked by avia
Find the solution(s) of the logarithmic equation
ln(x+4)+ln(x−4)=0
ln(x+4)+ln(x−4)=0
Answers
Answered by
Reiny
use rules of logs
ln(x+4)+ln(x−4)=0 , x > 4
ln( (x+4)(x-4) ) = 0
e^0 = (x+4)(x-4)
1 =x^2 - 16
x^2 = 17
x = ±√17 , but x > 4
so x = √17
ln(x+4)+ln(x−4)=0 , x > 4
ln( (x+4)(x-4) ) = 0
e^0 = (x+4)(x-4)
1 =x^2 - 16
x^2 = 17
x = ±√17 , but x > 4
so x = √17
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