# Homework Help: Adding negative exponents

1. Sep 17, 2009

### ckolin

1. The problem statement, all variables and given/known data

1/ 2^10 + 1/ 2^11 + 1/ 2^12 + 1 / 2^12 = ?

2. Relevant equations

3. The attempt at a solution

i am very confused with this problem as i thought that i would convert the 1/2^X numbers to 2^-X and then add the numbers together. The answer would be 1/2^45. and i know that isnt right. please help!
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Sep 17, 2009

### lanedance

multiply the whole expression (2^12)/(2^12) = 1

then keep the denominator as 2^12 and simplfy the numerator

3. Sep 17, 2009

### Staff: Mentor

You can write this sum as 2-10 + 2-11 + 2-12 + 2-12. (Is the last one supposed to be the same as the third one?)

Now, factor 2-12 out of each term (or 2-13 if the last term is 1/213).

Exponents add when you are multiplying factors, not when you are adding terms, so for example, it is not true that 1/22 + 1/23 = 1/25. Think about it: on the left you have 1/4 + 1/8 = 3/8. On the right, you have 1/32, which is nowhere near 3/8.

4. Sep 17, 2009

### ckolin

Ok i understand that 1/ 2^10 = 2^-10. But with the factoring wouldnt you need to factor out 2^-10? Can you show me the step by step solution. The answer is supposed to be 1/ 2^9

5. Sep 17, 2009

### mathie.girl

If you were going to factor the expression $$x^5 + x^4 + x^2$$, what would you factor out? The $$x^2$$, right? That's because it has the smallest exponent. So for your expression, you want to factor out the power with the smallest exponent, which is -12. You could also factor out $$2^{-10}$$, just like above you could factor out $$x^3$$ in my example, but it would leave you with fractions rather than whole numbers.

As an example, if you have $$4^{-2} - 4^{-4}$$, then our smallest exponent is -4 and we get:

$$4^{-2} - 4^{-4} = 4^{-4}[4^{-2-(-4)} - 4^{-4-(-4)}] = 4^{-4}[4^2 - 1] = 4^{-4}[15] = \frac{15}{4^4}$$

The exponent subtraction works exactly the same way.

6. Sep 17, 2009

### mathie.girl

Actually, looking at this particular problem, there's an easier way to do it.

The last two terms you have are both $$\frac{1}{2^{12}}$$, so when you add them together, what do you get? What about when that's simplified? Will that work again?

((I'm leaving my other comment about the exponents because it's useful to know, even if it's not necessary for this problem))