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by DavidG@VeritasPrep » Sun Feb 12, 2017 10:55 am
Brent@GMATPrepNow wrote:Here's a question I just made up.
Difficulty level: 700ish
If 2^m = 5 and 5^(1/n) = 3, then (3^n)/2^(-2m) =

A) 1/25
B) 1/5
C) 5
D) 25
E) 125
Answer: E
If 2^m = 5 and 5^(1/n) = 3, we can substitute 2^m in place of 5 in the second equation to get 2^m^(1/n) = 3. Raise both sides to the n to get
2^m = 3^n.

We're looking for (3^n)/2^(-2m), so substitute 2^m in place of 3^n, to get 2^m/2^(-2m) = 2^3m, or (2^m)^3. If 2^m = 5, then (2^m)^3 = 5^3 = 125. Answer is E
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by Brent@GMATPrepNow » Sun Feb 12, 2017 1:06 pm
Brent@GMATPrepNow wrote:If 2^m = 5 and 5^(1/n) = 3, then (3^n)/2^(-2m) =

A) 1/25
B) 1/5
C) 5
D) 25
E) 125
Another approach:

Given: 2^m= 5
We need to determine the value of 2^(-2m)
So, take 2^m= 5 and raise both sides to the power of -2
We get: (2^m)^(-2) = 5^(-2)
Apply Power of a Power rule to get: 2^(-2m) = 5^(-2)


Given: 5^(1/n) = 3
We need to determine the value of 3^n
So, take 5^(1/n) = 3 and raise both sides to the power of n
We get: [5^(1/n)]^n = 3^n
Apply Power of a Power rule to get: 5^1 = 3^n

So, (3^n)/2^(-2m) = (5^1)/[(5^(-2)]
= 5^3
= 125

Answer: E
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by GMATGuruNY » Sun Feb 12, 2017 1:26 pm
Brent@GMATPrepNow wrote:Here's a question I just made up.
Difficulty level: 700ish
If 2^m = 5 and 5^(1/n) = 3, then (3^n)/2^(-2m) =

A) 1/25
B) 1/5
C) 5
D) 25
E) 125
An alternate approach is to BALLPARK.
Since the answer choices are very spread out, we can approximate with integer values.

Since 2² = 4, 2^m = 5 implies that m ≈ 2.
Since √5 ≈ 2.24, 5^(1/n) = 3 implies that n ≈ 2.
Thus:
(3^n)/2^(-2m) ≈ (3²)/2¯� = (9)(16) = 144.
Only E is viable.

The correct answer is E.
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by Matt@VeritasPrep » Fri Feb 17, 2017 1:11 am
Recognizing the relationship between what you want and what you're given is crucial here.

We want 3� and 2�²�. We can get the first by raising 5⅟� to the n, and the second by raising 2� to the -2. Then we're set!
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