# Is this correct or should I use logs?

• May 24th 2011, 04:12 AM
Neverquit
Is this correct or should I use logs?
Hi did I simplify this correctly?
(2^x+2^(x-1)) / (2^x+2^(x+1))
Same base use multiplication rule

2^x+x-1 / 2^x+x+1
2^2x-1 / 2^2x+1
Same base so use division rule
2 ^ 2x-1 – (2x+1)
2 ^ 2x-1 – 2x-1
2 ^ -2
=1/4
• May 24th 2011, 04:20 AM
HallsofIvy
It is hard to tell what that is supposed to be without more parentheses. My first guess was that you meant
(2^x+ 2^(x-1))/(2^x+ 2^(x-1)) but that is equal to 1 because the numerator and denominator are the same.

But then by the third step that "-1" in the denominator has mysteriously become "+1". Did you intend
(2^x+ 2^(x-1))/(2^x+ 2^(x+1))?

In any case, if the problem is either of those, you can't use a "multiplication rule" because you don't have a multiplication.
It is true that (2^a)(2^b)= 2^(a+b) but you have 2^a+ 2^b, the sum, not the product.

If it really is [tex]\frac{2^x+ 2^{x-1}}{2^x+ 2^{x+1}}[tex], I recommend you factor $2^(x- 1)$ out of the numerator and $2^x$ out of the denominator. That will give
$\frac{2^{x-1}(2+ 1)}{2^x(1+ 2)}$
Simplify that.
• May 24th 2011, 04:30 AM
Neverquit
put brackets in