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07 Dec 2017, 21:22
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
65%
(hard)
Question Stats:
53% (02:10) correct
47%
(01:48)
wrong
based on 30
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If a^3 = 36, b^4 = 81, and c^5 = 125, which of the following must be true?
A. a>b>c
B. a>c>b
C. b>a>c
D. b>c>a
E. c>a>b
A. a>b>c
B. a>c>b
C. b>a>c
D. b>c>a
E. c>a>b
Archived Topic
Hi there,
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07 Dec 2017, 23:50
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Bunuel
Since \(a^3 = 36\), a is positive.
The value of a is in the range 3 < a < 4 because \(a^3 = 36\), which is between \(27(3^3) and 64(4^3)\)
Since \(b^4 = 81\), b could be both positive or negative. The value of b is 3.
Since \(c^5 = 125\), c is positive.
The value of c is in the range 2 < c < 3 because \(c^5 = 125\), which is between \(32(2^5) and 243(3^5)\)
Hence, a>b>c(Option A) must be true.
Bunuel
This question is a lot harder than it looks, IMO. I'd bet many will get lucky when choosing A.
If \(n^{odd} = positive\) number, \(n\) is positive
Approximate value of \(a\)
\(a^3 = 36\), and \(36\) is positive
\(a\) must be positive
The closest integer for \(a\) is \(3\), because
\(3^3 = 27\), so \(a\) must be a little bigger than \(3\)
\(a = 3+\)
Value of \(b\)
\(b^4 = 81\)
\(b = 3\) or \(b = -3\)
Approximate value of \(c\)
\(c^5 = 125\)
\(125\) is positive. \(c\) must be positive
The two integers closest to \(c\) are \(2\) and \(3\):
\(2^5=32\) and \(3^5=243\), where
\((2^5=32) < (c^5=125) < (3^5=243)\)
Compare \(c\) to positive \(b\)
\(c\) cannot be \(3\)
\(c^5=125 \neq{243}\) (not even close)
\(c \neq{3} \neq{+b}\)
\(c = 2+\)
Is \(b\) positive or negative?
If \(b = +3\)
\(a > b > c\) (Answer A)
If \(b = -3\)
\(a > c > b\) (Answer B)
Is \(b\) positive or negative? A quandary, maybe. The principal root rule is no help.*
Negative numbers raised to even powers: brackets
I think the key is the "bracket" rule for negative numbers raised to even powers. Think of "negative four, squared."
IF \(-4\) has brackets, then:
\((-4)^2 = 16\)
That means: \((-4)(-4) = 16\)
BUT IF \(-4\) has no brackets, then:
\(-4^2 = -16\)
That means: \(- [(4)(4)] = -16\)
Expanding: \(-4^2 = ?\)
By PEMDAS order of operations, we take exponents first.
So first we raise \(4\) to the second power: \(4^2 = 16\)
Then we take the negative of the result: \(-16\)
\(-4^2 = -16\)
\(b^4 = 81\)
IF \(b\) were negative, in order to equal positive 81, \(b\) would REQUIRE brackets. Per above:
\((-3)^4 = 81\) BUT
\(-3^4 = -81\)
There are no brackets around \(b\).
So \(b\) in \(b^4 = 81\) MUST be positive
\(b = 3\), and
\(a = 3+\)
\(c = 2+\)
\(a > b > c\)
ANSWER A
Please correct me if I am mistaken. I see no other way to choose A over B. The absolute value of \(b = 3\), but I do not see how absolute value tells us the sign of \(b\).
*That is
IF we are GIVEN
\(\sqrt[4]{b^4}\) , the answer \(b\), is positive. The radical sign itself tells us to take the principal fourth root, the positive one.
But we are given an exponent, not a radical sign, such that \(b^4 = 81\) has two solutions: \(3\) and \(-3\). In other words, given \(b^4 = 81\), there is no radical sign to indicate "Take the principal (positive) fourth root." Without such direction, we still do not know if \(b\) is positive or negative.
Use the "negative number raised to even power without brackets" rule: no brackets and a positive result? The number b is positive. 
