Gurpinder wrote:thx guys!

Good question.. Surprizingly, i havent seen this question in my 4 attempts of GMAT Prep yet.. Good to know their Question Bank is huge
So, the question asks for the value of x.
You want your LHS to be in the format 2^13, So I looked at the answer choices to see, which answer choice will give me:
x-2 = 13, solve for x = 15.
So your LHS is now:
2^13(2^2 - 2^0) = 3(2^13)
2^13(4-1) = 3(2^13)
2^13(3) = 3(2^13).

Hope this helps.

Hey guys,

Terrific question...I've always really liked this one and its derivatives. Again, with exponents I always come back to three guiding principles:

1) Multiply, and factor out addition/subtraction to enable yourself to multiply.

2) Find common bases, which usually means breaking down bases to prime factors.

3) Find patterns.

Either the first or the third principle will work here (they're all already in prime form so #2 doesn't really apply). To factor-and-multiply, it's important to note that you need to factor a common term. To get that term, it's a little tricky, but you can express:

2^x - 2^(x-2)

as:

2^(x-2) * 2^2 - 2^(x-2) to get that common term. It's a little bit backward (usually you add the exponents when multiplying the same base to simplify the term) but if we break that 2^x into two parts, we can replicate the 2^(x-2) as a common term to factor:

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

Now we have that 2^2 - 1 term, which we can calculate pretty quickly to be 3, and we then have:

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

You can divide both sides by 3, and now you have the same base set equal to two different exponents:

2^(x-2) = 2^13

So you can set the exponents equal to find that:

x-2 = 13
x = 15



One other way to do this (and the way I solved it when I first saw this question from scratch and had to formulate an explanation in <2 minutes) is to use that third principle and establish patterns with small numbers (which is often possible with exponent problems that use big numbers like 2^13).

I'd take x = 3 and see if I can get a 3*2^something term on the right out of it:

2^3 - 2^1 = 8 - 2 = 6. 6 = 3(2)

So I know that, even with really small numbers, I can use the 2^x - 2^(x-2) form to get something on the right that looks like 3(2^13). Now I'll try that with x = 4 and x = 5:

x = 4

2^4 - 2^2 = 16 - 4 = 12. 12 = 3(2^2)


x = 5

2^5 - 2^3 = 32 - 8 = 24. 24 = 3(2^3)

As I do this, I realize that the right hand side of the equation always ends up with 3 * 2^(x-2)...I always replicate the 3, and the 2 term is always the same as the x-2 exponent. So I can extrapolate that pattern to realize that 2^13 is going to be the same as 2^(x-2), making x 15.

thank you both!

this was an easy one

Ok, let me try.

So 2^x - 2^x-2 = 3(2^13).

2^x-2 can be simplified as 2^x divided by 2^2.
So the equation will become 2^x-(2^x)/4 = 3(2^13)

Solving Further, 4(2^x)-2^x=4*3*2^13(=3*2^15)

so, 2^x(4-1)=3*2^15

Thus x=15

Gurpinder wrote:thx guys!

I like such one-liners very much

2^x - 2^ (x - 2) = (4 - 1) × 2^13 = (2^2 - 2^0) × 2^13 = 2^15 - 2^13, x is [spoiler]15 on compare


Again D

Is it a D Day today, I found 3 Ds on a trot this morning[/spoiler]