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Non Integer Exponents??

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Source: — Data Sufficiency |
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by GMATGuruNY » Tue Sep 25, 2012 6:40 pm
dellaboemia wrote:If 3^a4^b = c, what is the value of b?

(1) 5^a = 25

(2) c = 36

ANSWER :C
SOURCE: KAPLAN QUESTION BANK
Statement 1: 5^a = 25
Since 5² = 25, a=2.
Thus, 3^a = 3² = 9.
Substituting 3^a = 9 into 3^a4^b = c, we get:
9(4^b) = c.
No way to solve for b.
INSUFFICIENT.

Statement 2: c=36
Substituting c=36 into 3^a4^b = c, we get:
3^a4^b = 36.
No way to solve for b.
INSUFFICIENT.

Statements 1 and 2 combined:
Substituting 3^a = 9 and c=36 into 3^a4^b = c, we get:
9(4^b) = 36
4^b = 4.
Since 4¹ = 4, b=1.
SUFFICIENT.

The correct answer is C.
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by anuprajan5 » Tue Sep 25, 2012 11:00 pm
GMATGuruNY wrote:
dellaboemia wrote:If 3^a4^b = c, what is the value of b?

(1) 5^a = 25

(2) c = 36

ANSWER :C
SOURCE: KAPLAN QUESTION BANK
Statement 2: c=36
Substituting c=36 into 3^a4^b = c, we get:
3^a4^b = 36.
No way to solve for b.
INSUFFICIENT.

The correct answer is C.
Mitch,

I have a question. If we prime factorize 36, we get 3^2 * 2^2. Wouldn't we able to determine what b is if we set in the format of 3^2*4^b

Regards
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by GMATGuruNY » Wed Sep 26, 2012 5:19 am
anuprajan5 wrote:
GMATGuruNY wrote:
dellaboemia wrote:If 3^a4^b = c, what is the value of b?

(1) 5^a = 25

(2) c = 36

ANSWER :C
SOURCE: KAPLAN QUESTION BANK
Statement 2: c=36
Substituting c=36 into 3^a4^b = c, we get:
3^a4^b = 36.
No way to solve for b.
INSUFFICIENT.

The correct answer is C.
Mitch,

I have a question. If we prime factorize 36, we get 3^2 * 2^2. Wouldn't we able to determine what b is if we set in the format of 3^2*4^b

Regards
Anup
It is important to recognize how a problem is restricted and how it ISN'T.
This DS places NO RESTRICTIONS on a and b: they can be integers, non-integers, positive, negative, etc.
Thus, there are an INFINITE number of solutions for (3^a)(4^b) = 36.
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by dellaboemia » Wed Sep 26, 2012 2:52 pm
Anup,

The answer is C. I actually had the same question that you had regarding factoring of 36. I guess it's a lesson in being rigorous about constraints.

Thanks Mitch!!
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