If 2^x - 2^x-2 = 3*(2)^13, What is the value of x?

This is from the GMAT Prep 1. Can someone please walk me through this?

LHS:
2^x(1- 1/4) = 2^x(3/2^2) = 3*2^(x-4). Now equate to RHS 3*2^(x-4) = 3*(2)^13, which means x-4 = 13, x=17.

-Raj.

can you explain how you converted 2^x - 2^x-2 into 2^x(1- 1/4)?

Hi Raj,

The correct answer is 15 not 17.

Hi MIT_Aspirant,

This is how I approached this prob.

2^(X) – 2^(X-2) = 3* 2^(13)
Taking 2^(X) common on the left side, we get:
2^(X)[1- 2^(-2)] = 3* 2^(13)
2^(X)[1- (1/4)] = 3* 2^(13) …Since 2^(-2)=(1/4)
2^(X)[(3/4)] = 3* 2^(13)
2^(X) = 3* 2^(13) * (4/3)
2^(X) = 2^(13+2) …Since we cancel 3’s and 4=2^(2)
X=15

Thanks,
Anand

Sorry about that.. 2^x(3/2^2) = 3*2^(x-4) is NOT right. It should say 2^x(3/2^2) = 3*2^(x-2), hence x-2 =17, x = 15.

I hope not to make such mistakes in the actual exam !!
-Raj.

ok guys, your opener works here, but there's an easier way.
remember - you should make connections between similar-looking problems. and this problem is similar to just about any other problem in which you're factoring a common power out of a polynomial.

here's the deal: when you factor a power out of a polynomial, which power do you factor out: the smallest common power, or the biggest common power?
that's right, the smallest one. if you have x^5 - x^2, you only factor out x^2, not x^5.
analogy:
on this problem, just factor out 2^(x - 2).
this gives
left hand side = [2^(x - 2)](2^2 - 1)
= [2^(x - 2)](3)
you can then cancel the 3's, leaving 2^(x - 2) = 2^13. therefore, x - 2 = 13, so x = 15.
done.

btw, the most important part of this post is the 'make analogies' part. if you focus on the parts of each problem that remind you of other problems, then you're going to face a lot less memorization and a lot more progress.

Instead of focusing on the LHS , we can try and simplify the RHS

LHS = 3 * 2^13

can be written as (2^2 - 1) * 2^13.

= 2^15 - 2^13.

now RHS = 2^x - 2^(x-2) = LHS = 2^15 - 2^13

therefore x = 15.

this is pretty neat, although it's not algorithmic; i.e., it's a nice little heuristic that will shave some time off the problem if it happens to work, but there's no guarantee that it will work. in other situations, it could crash and burn.
on this particular problem, though, it's beautiful.

factoring approaches, on the other hand, are usually algorithmic; i.e., they'll work, regardless of the nature of the problem.