natural logs?

Question

ln(x+5)=ln(x-1)-ln(x+1)



i dont know why im having soo much trouble with this one...usually these come easy to me....

help please!!!

Answer 1

ln(x+5)=ln(x-1)-ln(x+1)
In(X+5)=In((x-1)/(x+1))
x+5=(x-1)/(x+1)
(x+5)*(x+1)=x-1
x^2+6x+5=x-1
x^2+5x+6=0
(x+3)(x+2)=0
x=-3,-2

Answer 2

It doesn't matter what base is used in the logs, the operations are the same. So.....
First, combine the terms on the right to get
ln(x+5)= ln([x-1]/[x+1])
Next take both sides as exponents of "e"
x+5 = [x-1]/[x+1]
Multiply both sides by x+1, combine terms on the left side. Then subtract {x-1} from both sides. Again combine terms on the left side. See if the quadratic form is factorable (probably it won't), and solve for roots.

Answer 3

lnA + ln B = ln(AB)
lnA - ln B = ln(A/B)

ln(x+5)=ln(x-1)-ln(x+1)
ln(x+5)=ln(x-1)(x+1)
(x+5) = (x-1)(x+1) ------- as long as (x+5), (x-1) and (x+1) are all > 0
x + 5 = x^2 - 1
x^2 - x - 6 = 0
x^2 - 3x + 2x - 6 = 0
x(x-3)+2(x-3) = 0
(x-3)(x+2) = 0
x = {-2, 3}

(x+5), (x-1) and (x+1) must all be >0.
This disallows x = -2 as a solution
x = 3