Simplifying Negative Exponents: How to Add Fractions with Negative Exponents

[ckolin]

Homework Statement

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

Homework Equations

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 isn't right. please help!

 

[lanedance]

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

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

 

[Mark44]

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.

 

[ckolin]

Ok i understand that 1/ 2^10 = 2^-10. But with the factoring wouldn't 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

 

[mathie.girl]

Ok i understand that 1/ 2^10 = 2^-10. But with the factoring wouldn't 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

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.

 

[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))