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

Step by step:
\(2^x-2^{x-2}=3*2^{13}\)

\(2^x-2^{x}*2^{-2}=3*2^{13}\)

\(2^x-\frac{2^{x}}{2^{2}}=3*2^{13}\)

\(2^x(1-\frac{1}{2^{2}})=3*2^{13}\)

\(2^x(\frac{2^2-1}{2^{2}})=3*2^{13}\)

\(2^x*\frac{3}{2^{2}}=3*2^{13}\)

\(2^x=2^{15}\)

\(x=15\).

Hope it's clear.

Just another way of doing this sum:
\(2^x-2^{x-2}= (4-1)*2^{13}\)

or \(2^x-2^{x-2} = 2^{15}-2^{13}\)

Thus, x = 15.

I can explain that:

\(2^x-2^x^-^2=2^x(1-2^-^2)=2^x(1-\frac{1}{2^2})=\frac{2^x*3}{4}=3*2^1^3\)
divide by 3
\(\frac{2^x}{4}=2^1^3=2^x^-^2=2^1^3\)

Hope now it's clear

Hey ! Unfortunately, I still don't get where that '3' comes from... Just don't see it... Could anyone explain further in more detail?! Would be great!

Hey, now its clear where the 3 comes from but how did u get the 2^2 there? Sry, was a long day... I bet its easy as hell but I'm just confused atm..

I think you should brush up fundamentals:

\(2^{-2}=\frac{1}{2^2}\).

Hey, that I understand... but how did u get the 2^2 onto the fraction again to get 2^2-1/2^2 ? I must miss something very fundamental indeed..

Can you please tell me what didn't you understand below?

\(2^x-2^{x-2}=3*2^{13}\)

\(2^x-2^{x}*2^{-2}=3*2^{13}\)

\(2^x-\frac{2^{x}}{2^{2}}=3*2^{13}\)

Hey, sry for not getting it. In the extract shown above I understand everything.

What I seem to not get is how you got from 2^x (1-1/2^2) to 2^x (2^2-1/2^2).

Oh wait... did u just multiply with the common denominator (2^2*(-1)) ?

\(1-\frac{1}{2^{2}}\) --> \(\frac{2^2}{2^2}-\frac{1}{2^{2}}\) --> \(\frac{2^2-1}{2^{2}}\)

\(2^x - 2^{x-2}=3*2^{13}\) can be written as..
\(2^x - 2^x/2^2=3*2^{13}\)..
\(2^x(1-1/4)=3*2^{13}\).
\(3* 2^{x-2}=3*2^13\)....
or x-2=13..
x=15
ans D

Hi All,

Factoring the equation is a useful approach here. You can actually take it a step further by TESTing THE ANSWERS (you might find it easier to manipulate numbers than to manipulate variables). One of those numbers IS the value of X, so you can plug them in, do the necessary math and find the one value that balances out the equation. Here's how to approach it:

Since the "right side" of the equation is greater than 2^14, X must be bigger than 14 (since the "left side" of the equation involves subtraction). So we can eliminate A, B and C.

Let's TEST answer D: 15

If X = 15, then...

2^15 - 2^13 can be factored into...
(2^13)(2^2 - 1) =
(2^13)(3)

This is EXACTLY what's on the "right side" of the equation, so X MUST be 15

Final Answer:
GMAT assassins aren't born, they're made,
Rich

We are given the equation 2^x – 2^(x-2) = 3(2^13) and we must determine x. We can start by rewriting the equation as follows:

2^x – (2^x)(2^-2) = 3(2^13)

Next we can factor out 2^x from the left hand side of the equation.

2^x(1 – 2^-2) = 3(2^13)

2^x(1 – 1/4) = 3(2^13)

2^x(3/4) = 3(2^13)

Multiplying both sides by 4/3 gives us:

2^x = 3(2^13) * 4/3

2^x = (2^13) * 4

2^x = 2^13 * 2^2

2^x = 2^15

x = 15

Answer: D

Hey guys I just want to know how we got rid of the subtraction sign and where did the 1 pop up from?
Everything else I understand just this step seems to be consolidated in the explanations.
2^x−2^x/2^2

2^x(1−1/2^2)

Factor out 2^x from \(2^x-\frac{2^{x}}{2^{2}}\) to get \(2^x(1-\frac{1}{2^{2}})\).

The same way after factoring m from m - m/4 you get m(1 - 1/4).

Hey guys,

This has always been a favorite problem of mine - the first time I saw it, an instructor had blanked on how to solve it and emailed me a photo from his phone asking for help. He had excused himself from a tutoring session and needed me to explain how to solve it (correctly, of course) within a few minutes, so the pressure was on!


Bunuel's explanation is perfect (as always), but when the pressure was on and I wasn't exactly thinking about factoring, I did this instead - I looked to see if there were a pattern in the subtraction at left (2 to an exponent minus 2 to another exponent, two less) that would always produce 3*something on the right. So I did:

x = 3 and x-2 = 1
2^3 - 2^1 = 8 - 2 = 6

And 6 = 3(2^1), so I had a start.


x = 4 and x-2 = 2
2^4 - 2^2 = 16 - 4 = 12

And 12 = 3(2^2), so the pattern held

x = 5 and x-2 = 3
2^5 - 2^3 = 32 - 8 = 24

And 24 = 3(2^3), and the pattern became clear...
The operation at left was always producing 3*2^(x-2) as its answer, so if x-2 = 13, then x = 15.


Strategically, using small numbers to establish patterns works pretty well when huge numbers (like 3(2^13)) are in play, and when exponents are involved (exponents are essentially just repetitive multiplication, so there are bound to be some repetitive patterns involved). If you can factor like Bunuel did, that's a great way to go...but I'd recommend having the "prove patterns w/ small numers" ideology in your arsenal!

Bunuel do you have any similar problems?