Large exponents

cooperbaker wrote:If 2^x - 2^(x-2) = 3(2^13), what is the value of x?

a) 9
b) 11
c) 13
d) 15
e) 17


OA: D

I didn't know how to solve this question using the properties of exponents. What I did realize is I could subtract 10 from all the exponents then use real numbers in my head -- (i.e. 3(2^3) = 24)

Is this a good way to go about it, or is this wasting time? My $40 Kaplan book doesn't explain what to do when adding or subtracting large exponents with the same base.... :?

Your method here is clever, just as long as you add 10 back at the end. I'd be a bit concerned that you'd have trouble doing the same with similar problems, however. There are a few ways to solve the equation as provided; truplayer gives one good approach above, and I'll suggest another. In general, when you're adding or subtracting terms with the same base and different powers, you'll want to factor out the term with the smallest power. For example, if you see x^5 - x^3, in most cases it will be useful to factor out x^3, to get:

x^5 - x^3 = x^3(x^2 - 1)

(and depending on the question, you might want to factor the x^2 - 1 = (x+1)(x-1) as well). In the question in the original post, the smaller power is (x-2), so we can factor out 2^(x-2), to get:

2^x - 2^(x-2) = 3(2^13)
(2^(x-2))*(2^2 - 1) = 3(2^13)
(2^(x-2))*3 = 3(2^13)
2^(x-2) = 2^13
x-2 = 13
x = 15

awilhelm wrote:Could anyone please explain why my approach doesn't lead to the correct answer?

2^x - 2^(x-2) = 3(2^13)
2^x - 2^x * 2^-2 = 3(2^13)
2^x(1 - 2^-2) = 3(2^13)
2^x(1 - 1/4) = 3(2^13)
2^x(3/4) = 3(2^13)
4 * 2^x * 3 = 3(2^13) * 4
12 * 2^x = 12 * 2^13
2^x = 2^13
x = 13

It's almost entirely right, and it will lead to the right answer if you fix one small error. In the line highlighted in red above, you intended to multiply both sides by 4, but on the left side, you've actually multiplied by 16 -- you cancelled the 4 in '3/4' but you've introduced an additional 4 as well.