ln(x+3) + ln(x-1) =0?

Question

also ln(x+2) - ln(x+1) = 1

Answer 1

ln (a) + ln(b) = ln (ab)

so

ln(x+3) + ln(x-1) =0 ==>
ln [ (x+3) (x-1) ] = 0 ==>
e^ln [ (x+3) (x-1) ] = e^0 ==>
(x+3)(x-1) = 1
x^2 + 2x - 4 = 0

x = -2 (+/-) sqrt (20) / 2

x1 = -1 - sqrt (5)
x2 = -1 + sqrt (5)

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ln(x+2) - ln(x+1) = 1 is similar:

ln [(x+2) / (x+1)] = 1
(x+2) / (x+1) = e
x+2=e*x+e
x(e-1)=2-e
x=(2-e)/(e-1)

Answer 2

ln [(x + 3).(x - 1)] = 0
(x + 3).(x - 1) = e^(0) = 1
x² + 2x - 4 = 0
x = [- 2 ± 2√5] / 2
x = - 1 ± √5

ln (x + 2) / (x + 1) = 1
(x + 2) / (x + 1) = e
x + 2 = e x + e
(1 - e).x = e - 2
x = (e - 2) / (1 - e)