SOLUTION: Solve a)log(base 2)(x + 1)+ log(base 2)(x-1) = 3 b)log(base 2)(x^2-1) = 3 c)explain why the equations in part a and b are not equivalent Can you please help me out? Thanks so

Question 867265: Solve
a)log(base 2)(x + 1)+ log(base 2)(x-1) = 3
b)log(base 2)(x^2-1) = 3
c)explain why the equations in part a and b are not equivalent
Can you please help me out? Thanks so much in advance:)
Answer by edjones(8007) (Show Source): You can put this solution on YOUR website!
a)
log(base 2)(x + 1)+ log(base 2)(x-1) = 3
2^log(base 2)(x + 1)*2^log(base 2)(x-1)=2^3
(x+1)(x-1)=8
x^2=9
x=3
.
b)
log(base 2)(x^2-1) = 3
2^log(base 2)(x^2-1) =2^3
x^2-1=8
x^2=9
x=3, x=-3
.
c) x can't equal -1 in a) because a log cannot equal 0 or be negative.
.
Ed

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