The issue here is that inside statement
2^{4n} = k^n, I think it is implicit that this is true for all
n, or else it would say something like “n is some integer for which the following is true”.
A student here makes the argument that k=-16 also satisfies, and Greg changes his answer.
However, suppose k=-16, take n=1:
2^4 = 16 \neq (-16)^1.
The only answer that solves this equation for all n is 16.
2^(4k) = (-16)^k - Wolfram|Alpha As you can see, it is only “True” only for even integer power n.
The other student who gives k=\textit{whatever} if n=0. This is only true for a single case of n. k can’t be whatever in general.
I think if this were the desired question, it should say “n is some integer for which the following is true”, otherwise I assume “for all n”, and answer A would be correct, as Greg initially said. Thoughts?I don’t think ETS would do something so ambiguous, right?
Thanks.
This is correct, but there is no indication that we must be looking for a value of k that solves the equation no matter what n is.
Take a step back. The question is asking “for what k is the equation true”. Crucially, it does not tell anything about that n could be, which means that we have to assume that it can be any real number. As a result, we have to check whether k = 16 no matter what n is. This isn’t the case - as people in the video pointed out, if n = 0, k can be anything. As a result, it isn’t always true that Quantity A is greater than Quantity B, making the correct option D.
Noticing these traps and shortcuts in a QC problem is pretty important, as that’s one way to finish these questions quickly. This is what Greg calls the “equal-not equal” strategy.
The mistake we made with this question was that we forgot to set constraints that would stop people from “short-circuiting” the question to get D, unexpectedly making the question hard for a Level 1 student (and causing us embarrassment) since the majority didn’t spot the edge cases. For example, by saying something like “n is an odd positive integer”.
