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03 Feb 2015, 09:57
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For integers a and b, 16a = 32^b. Which of the following correctly expresses a in terms of b?
A. a = 2^b
B. a = 4^b
C. a = 2^(5b − 4)
D. a = 4^(5b − 4)
E. a = 2^(5b)
Kudos for a correct solution.
A. a = 2^b
B. a = 4^b
C. a = 2^(5b − 4)
D. a = 4^(5b − 4)
E. a = 2^(5b)
Kudos for a correct solution.
03 Feb 2015, 10:40
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Bunuel
For integers a and b, 16a = 32^b. Which of the following correctly expresses a in terms of b?
A. a = 2^b
B. a = 4^b
C. a = 2^(5b − 4)
D. a = 4^(5b − 4)
E. a = 2^(5b)
Kudos for a correct solution.
\(16a = 32^b\)A. a = 2^b
B. a = 4^b
C. a = 2^(5b − 4)
D. a = 4^(5b − 4)
E. a = 2^(5b)
Kudos for a correct solution.
\(2^4 a = (2^5)^b\)
a = (2)^(5b-4)
[C]
03 Feb 2015, 10:56
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I tried this one with an example, although I am not quite sure if this is correct. I would appreciate if there would be a common rule for this.
I tried it out with b = 2 and only c correctly expresses the following:
16a = 32^b b = 2
16a = 1024
a = 64
choice c --> a = 2^6 = 64
However, this is a quite time consuming process. Is there something faster?
Answer C
I tried it out with b = 2 and only c correctly expresses the following:
16a = 32^b b = 2
16a = 1024
a = 64
choice c --> a = 2^6 = 64
However, this is a quite time consuming process. Is there something faster?
Answer C
