Solve this by factoring out 2^(x-2) out of the equation to make it look like the other side of the equation:

2^x - 2^(x-2) = 3*(2^13)
2^(x-2)*(2^2 - 1) = 3*(2^13)
2^(x-2)*(4 - 1) = 3*(2^13)
2^(x-2)*(3) = 3*(2^13)
2^(x-2) = (2^13)
x = 15

If that seems somewhat confusing to you, think of it with variables. If you have a side of an equation that is y^15 - y^13 (note how this is the same form with one exponent is x and the other exponent is x-2). You can factor this equation out to show y^13*(y^2 - 1). In this case y = 2, so the inside of the parenthesis simplifies to 3.

In pretty much all GMAT problem solving questions I've seen that are of this format, you are going to be able to simplify the complicated side of the equation to match the easier looking side of the equation. So start with that in mind and work from there.

thank you so much, but i dont understand this step in your above solving...


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

how does 2 ^ X turn into 2 ^ 2-1 ? I've never seen anyone do that before.

You are factoring out 2^(x-2). If you break it down by each part, (2^(x-2))*x^2 = 2^x as you just add the exponents together when you are multiplying (x-2+2 = x). When you factor out 2^(x-2) out of itself, it factors to 1. Does that make sense?

Hmm, i thought 2^ X-2 was 2^X/2^2

The whole point of this is manipulate the left side of the equation to look like the right side of the equation.
This is purely exponential equation. You have to know the exponential
Rules by heart in order to do this. This website can help. It helped me get down exponential rules.

https://www.ltcconline.net/greenl/course ... intexp.htm

Mclaughlin wrote:Hmm, i thought 2^ X-2 was 2^X/2^2

It is. Wawatan solved the problem using that as a method and that is totally right too. I just factored out 2^(x-2). I always found that I understood it better when I thought of it in variables (which is usually backwards).

2^x - 2^(x-2) is of the same format as y^x - y^(x-2)

So the equations:

y^15 - y^13
y^10 - y^8
y^3 - y

are all in that same format and can be simplified down to:

y^13*(y^2-1)
y^8*(y^2-1)
y*(y^2-1)

Notice how all the equations have the same format of:

y^(x-2)*(y^2-1)

It's the same thing in this equation, only y=2. So plugging that into the equation is:

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

There is more than one way to simplify this as wawatan showed, so definitely use what is easier to you. I just find this easier for myself.

I hope this helps! Good luck

wawatan,

I'd really be interested in seeing the link. However, the link above does not work, at least on my computer.

hi alex,
thanks for reminding me! i just edited my post. now it should work!!!

Kool, thanks. Though it doesn't contain examples as complex as the one above. I have found these kinds of problems the most challenging. 8 times out of 10 I get them wrong. I guess it's a bit late to start practicing them now.

hey alex,
i totally understand what you mean. the reason i know how to do this problem is that i took a precalculus course in college and high school so i kinda remember exponents really really well b/c i used to do bad in those. these types of problems are actually coming from a precalculus college book. i don't know the exact name of the book but if you check out any precalculus college book they will have samples of these problems. i do agree though. this is the more complex exponent problem.

I have another way to solve this problem

Backsolving

lets take x=15 from the answer choice

2^15-2^15-2= 3(2^13

now

2^15-2^13=3(2^13)

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

2^13(4-1) =3(2^13)


2^13(3)=3(2^13)

I have another way to solve this problem

Backsolving

lets take x=15 from the answer choice

2^15-2^15-2= 3(2^13

now

2^15-2^13=3(2^13)

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

2^13(4-1) =3(2^13)


2^13(3)=3(2^13)

you can definitely try backsolving however realize that it will be time consuming. it will not be the case that you would find the right answer right away, so you would have to plug in numbers from all 5 answer choices to find the correct answer.

however lets say you did algebra to find the correct answer and you are not sure if your answer is correct, then you should plug in x to make sure that your algebra is done correctly.