If n is a positive integer, is the value of b - a at least twice the value of 3^n - 2^n?
(1) a= 2^(n+1) and b= 3^(n+1)
(2) n = 3
(1) a= 2^(n+1) and b= 3^(n+1)
(2) n = 3
Thanks
Airan
Airan
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If n is a positive integer, is the value of b - a at least
Source: Beat The GMAT — Data Sufficiency |
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Ian Stewart
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Looking at Statement 1 alone:
The question is asking:
is 3^(n+1) - 2^(n+1) > 2*(3^n - 2^n) ?
i.e. is
3*3^n - 2*2^n > 2*3^n - 2*2^n
or, equivalently, is
3*3^n > 2*3^n
which clearly must be true, since 3^n must be positive. So the first statement is sufficient.
edit: technically, all of the inequalities above really should be 'greater than or equal to', because the question says 'at least', not 'greater than', but this does not affect the solution.
The question is asking:
is 3^(n+1) - 2^(n+1) > 2*(3^n - 2^n) ?
i.e. is
3*3^n - 2*2^n > 2*3^n - 2*2^n
or, equivalently, is
3*3^n > 2*3^n
which clearly must be true, since 3^n must be positive. So the first statement is sufficient.
edit: technically, all of the inequalities above really should be 'greater than or equal to', because the question says 'at least', not 'greater than', but this does not affect the solution.
Need a little help..
Is b-a greater than twice the value of 3^n - 2^n?
b - a > 2(3^n - 2^n)
b - a > 6^n - 4^n
Statement I:
b = 3^(n+1) a = 2^(n+1)
Added to Q Stem:
3^(n+1) - 2^(n+1) > 6^n - 4^n ?
If n = 1, 9 - 4 > 6 - 4 YES
If n = 2, 27 - 8 > 36 - 16 NO
.. I think I've been studying a bit too long today. What the heck am I missing?
Is b-a greater than twice the value of 3^n - 2^n?
b - a > 2(3^n - 2^n)
b - a > 6^n - 4^n
Statement I:
b = 3^(n+1) a = 2^(n+1)
Added to Q Stem:
3^(n+1) - 2^(n+1) > 6^n - 4^n ?
If n = 1, 9 - 4 > 6 - 4 YES
If n = 2, 27 - 8 > 36 - 16 NO
.. I think I've been studying a bit too long today. What the heck am I missing?
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Brent@GMATPrepNow
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The problem is highlighted above in blue.ShaneK wrote:Need a little help..
Is b-a greater than twice the value of 3^n - 2^n?
b - a > 2(3^n - 2^n)
b - a > 6^n - 4^n
Statement I:
b = 3^(n+1) a = 2^(n+1)
Added to Q Stem:
3^(n+1) - 2^(n+1) > 6^n - 4^n ?
If n = 1, 9 - 4 > 6 - 4 YES
If n = 2, 27 - 8 > 36 - 16 NO
.. I think I've been studying a bit too long today. What the heck am I missing?
2(3^n - 2^n) does not equal 6^n - 4^n
To show that (2)(3^n) does not equal 6^n, try n=2
We get (2)(3^2) = 6^2
Evaluate to get: 18 = 36
Using similar techniques, we can also show that (2)(2^n) does not equal 4^n
Aside: (2)(2^n) = (2^1)(2^n) = 2^(n+1)
Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
I was 98% sure you couldn't, but tested it and made it work- I made a mental math error in my "check".
I had
If n = 2, 2(3^n - 2^n) = 2(5) = 10
6^2 - 4^2 = 36 - 16 = 20
After I did this second round, I was like "oh, I forgot to x2 the first one" and essentially multiplied by 4.
Side note: If you double the coefficient of an exponent, are you essentially quadrupling the answer?
i.e. 3^2 = 9, 6^2 = 36.. this makes sense, because by doubling, you've added 2^n.. so you're not always quadrupling, you're 2^n power'ing, because 6^n is 2^n(3^n).
I know the rules, but exponents always mess with my brain!
Thanks for your help Brent
I had
If n = 2, 2(3^n - 2^n) = 2(5) = 10
6^2 - 4^2 = 36 - 16 = 20
After I did this second round, I was like "oh, I forgot to x2 the first one" and essentially multiplied by 4.
Side note: If you double the coefficient of an exponent, are you essentially quadrupling the answer?
i.e. 3^2 = 9, 6^2 = 36.. this makes sense, because by doubling, you've added 2^n.. so you're not always quadrupling, you're 2^n power'ing, because 6^n is 2^n(3^n).
I know the rules, but exponents always mess with my brain!
Thanks for your help Brent
GMAT/MBA Expert
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Brent@GMATPrepNow
- GMAT Instructor
- Posts: 16207
- Location: Vancouver, BC
- Thanked: 5254 times
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Hey Shane,ShaneK wrote: Side note: If you double the coefficient of an exponent, are you essentially quadrupling the answer?
i.e. 3^2 = 9, 6^2 = 36.. this makes sense, because by doubling, you've added 2^n.. so you're not always quadrupling, you're 2^n power'ing, because 6^n is 2^n(3^n).
Your question does not match your example.
Based on your example, I think your question is, "If you double the base are you essentially quadrupling the answer?"
In other words, "Is (2x)^n four times the value of x^n?"
The answer here is no. It's only true when n = 2.
Let's take a closer look at (2x)^n:
(2x)^n = [(2)(x)]^n = [(2)^n][(x)^n]
This tells us that (2x)^n is (2)^n times the value of (x)^n
So, if n=2, (as it did in your example) then (2)^n = 4, so the value of the power is, indeed, quadrupled.
However, if n does not equal 2, then the value is not quadrupled.
Example: compare 3^1 with 6^1.
We've doubled the base, but the value is not quadrupled.
Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
There a quicker way to get the right answer on this problem, even if you're not sure about the math involved. Our question asks us about the variables b, a, and n (relative to powers of 3 and 2). We need information about each of these variables to answer the question.
Start with the easy statement - statement (2). This gives us a value for n, but tells us nothing about b or a, so it's clearly insufficient. We can rule out B and D as answer choices.
If you weren't sure what to do with statement (1) alone, consider this: if we were to put the statements together, we already have a value for n, so we could clearly solve to get values for a and b. So clearly the statements together are sufficient, but... did we do any conceptual work? We just plugged values and got values out. We didn't do any thinking about the idea of exponents.
If it's a little too obvious that the statements together will solve it, then C IS A TRAP! We've eliminated B and D, and E is impossible (clearly it's solvable). So if C is a trap, the answer must be A! You don't really even need to do the algebra and prove it to yourself. (When you're studying, of course you want to understand why the algebra works. But when you're taking the test, don't stop to prove it! A is the only reasonable answer choice. Guess it and move on).
Here's some more info on common DS traps: https://www.manhattangmat.com/blog/index ... ncy-traps/
Start with the easy statement - statement (2). This gives us a value for n, but tells us nothing about b or a, so it's clearly insufficient. We can rule out B and D as answer choices.
If you weren't sure what to do with statement (1) alone, consider this: if we were to put the statements together, we already have a value for n, so we could clearly solve to get values for a and b. So clearly the statements together are sufficient, but... did we do any conceptual work? We just plugged values and got values out. We didn't do any thinking about the idea of exponents.
If it's a little too obvious that the statements together will solve it, then C IS A TRAP! We've eliminated B and D, and E is impossible (clearly it's solvable). So if C is a trap, the answer must be A! You don't really even need to do the algebra and prove it to yourself. (When you're studying, of course you want to understand why the algebra works. But when you're taking the test, don't stop to prove it! A is the only reasonable answer choice. Guess it and move on).
Here's some more info on common DS traps: https://www.manhattangmat.com/blog/index ... ncy-traps/
Ceilidh Erickson
EdM in Mind, Brain, and Education
Harvard Graduate School of Education
EdM in Mind, Brain, and Education
Harvard Graduate School of Education
