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If \(x^{a-b}=2\) and \(x^{a+b}=32,\) and if \(x, a,\) and \(b\) are greater than \(0,\) which of the following must be

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by Vincen » Thu Sep 03, 2020 6:10 am

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Difficulty—

If \(x^{a-b}=2\) and \(x^{a+b}=32,\) and if \(x, a,\) and \(b\) are greater than \(0,\) which of the following must be true?

I. \(a=3\)
II. \(x^b=4\)
III. \(b=\dfrac23a\)

A. I only
B. II only
C. I and II
D. II and III
E. I, II, and III

Answer: D

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Re: If \(x^{a-b}=2\) and \(x^{a+b}=32,\) and if \(x, a,\) and \(b\) are greater than \(0,\) which of the following must

by Scott@TargetTestPrep » Tue Sep 08, 2020 9:47 am
Vincen wrote: ↑
Thu Sep 03, 2020 6:10 am
 If \(x^{a-b}=2\) and \(x^{a+b}=32,\) and if \(x, a,\) and \(b\) are greater than \(0,\) which of the following must be true?

I. \(a=3\)
II. \(x^b=4\)
III. \(b=\dfrac23a\)

A. I only
B. II only
C. I and II
D. II and III
E. I, II, and III

Answer: D

Solution:

Dividing the second equation by the first, we have:

x^(2b) = 16

Since x and b are greater than 0, we take the square root and obtain:

x^b = 4

So, II is true.

Multiplying the original two equations, we have:

x^(2a) = 64

Since x and a are greater than 0, we take the square root and obtain:

x^a = 8

Since 8 = 4^(3/2), we have:

x^a = (x^b)^(3/2)

x^a = x^(3b/2)

Since x > 0 and x is obviously not 1, we cancel out x, and we have::

a = 3b/2

2a/3 = b

So, III is true also.

However, there is no way we can determine if a = 3. So, I might not be true.

Answer: D

Scott Woodbury-Stewart
Founder and CEO
[email protected]

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