log (x+3)+log(x-1)=log(x²-2x-3) is satisfied for ?

Question

A) all real values of x

B)no real values of x

C)all real values of x=0

D)no real of x, except x=0

E)all real values of x =1

Answer 1

Log (x+3) + log (x-1) = log (x^2 - 2x - 3)

log (x+3).(x-1) = log (x^2 -2x -3)

log (x^2 + 2x -3) = log (x^2 -2x -3)

x^2 + 2x -3 = x^2 -2x -3

2x = -2x

only true if x = 0

but remember arg of a log function must be > 0

if x = 0, log (x-1) = log (-1) is not defined

so no real values for x

Answer 2

I apologize for mis-copying the problem

log( (x+3)(x-1)) = log (x^2 - 2x - 3)

(x+3)(x-1) = x^2 - 2x - 3

x^2 -x + 3x - 3 = x^2 - 2x - 3

x^2 + 2x - 3 = x^2 - 2x - 3

2x = -2x

4x= 0

x=0

as Katherine pointed out, it does not check out

because of the negative arguments passed to the log.