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

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- May 24th 2011, 03:12 AMNeverquitIs 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, 03:20 AMHallsofIvy
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 $\displaystyle 2^(x- 1)$ out of the numerator and $\displaystyle 2^x$ out of the denominator. That will give

$\displaystyle \frac{2^{x-1}(2+ 1)}{2^x(1+ 2)}$

Simplify that. - May 24th 2011, 03:30 AMNeverquitput brackets in
Sorry added brackets

what I mean to state was 2 to the power of x plus 2 to the power of x minus one and that is all divided by 2 to the power of x plus 2 to the power of x plus 1.