If n is a positive integer, is the value of b - a at least

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.

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?

The problem is highlighted above in blue.

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

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).

Hey Shane,
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

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/