Help with this log problem! Please!!?

Question

It was on a quiz I just had and I can't figure out how to get the answer.

You just have to write it as one log... I get confused where I'm dividing one part of the log inside another difference. Please help.


2log base 2(x+1) - log base 2(x+3) - log base 2 (x-1)

Answer 1

log2(x+1)^2-log2(x+3)-log2(x-1)=
log2 (x+1)^2/(x+3)(x-1).
user theat:logA-logB=logA/B
logA+logB=logAB

Answer 2

Factor out the minus sign of the last two terms, and the answer becomes clear.

2log base 2 (x+1) - log base 2 (x+3) - log base 2 (x-1) is the same as:

2log base 2(x+1) - [log base 2 (x+3) + log base 2 (x-1)].

This implies that (x+3) and (x-1) are factors in the denominator of the expression.

Furthermore, 2log base 2 (x+1) = log base 2 [(x+1)^2].

So what we are left with is:

log base 2 [(x+1)^2/(x+3)(x-1)].

Answer 3

Answer is
2log(x+1) - ( log(x+3)(x-1))
log base 2 (x+1)^2/(x+3)(x-1)

Answer 4

2log base 2(x+1) - log base 2(x+3) - log base 2 (x-1)
=log base 2(x+1)² - log base 2(x+3) - log base 2 (x-1)
=log base 2{(x+1)²/[(x+3)(x-1)]}

Answer 5

Let log base 2 be denoted log_2:

2log_2(x+1) = log_2(x+1) + log_2(x+1) = log_2(x+1)^2

Let a = (x+1), b = (x+3), c = (x-1). Then, we have
log_2(a^2) - log_2(b) - log_2(c) = log_2[a^2/(b/c)]
= log_2[(a^2)(c)/b]
[Since 1/(b/c) = c/b]

Making the substitutions, yields:
log_2[[(x+1)^2 * (x-1)]/(x+3)]