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Difficulty:
35%
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Question Stats:
70% (01:25) correct
30%
(01:31)
wrong
based on 80
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If a and b are nonzero integers and \(a^{3b} = a^{b-6}\), what is the value of \(b^{a}\)?
(A) a^3 = 8
(B) a^2 = 4
(A) a^3 = 8
(B) a^2 = 4
I understand that (A) is sufficient. But if I simplify the given equation algebraically -
a^3b = a^b–6
a^3b/ a^b–6 = 1
a^(3b-b+6) = 1
a^(2b+6) = 1
(a^2)^b * a^6 = 1
(a^2)^b * (a^2)^3 = 1
This would make (B) sufficient, wouldn't it? Am I missing something in my simplification? Appreciate any pointers. Thanks.
a^3b = a^b–6
a^3b/ a^b–6 = 1
a^(3b-b+6) = 1
a^(2b+6) = 1
(a^2)^b * a^6 = 1
(a^2)^b * (a^2)^3 = 1
This would make (B) sufficient, wouldn't it? Am I missing something in my simplification? Appreciate any pointers. Thanks.
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17 Nov 2014, 01:35
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codensnap
\(a^{3b}\) = \(a^{b-6}\)
as bases are same on both side, thus 3b=b-6
or 2b=-6, b=-3
now to find the value of \(b^{a}\), we need to find the value of a.
st.1 \(a^{3}\)=8
\(a^{3}\)=\(2^{3}\). thus a must be 2. hence sufficient
st.2 \(a^{2}\)=4
both a=2, and a=-2 can satisfy the above equation. hence st.2 alone is not sufficient.
17 Nov 2014, 02:58
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Given that a^3b=a^(b-6)
a^(3b-b+6)=1
2b+6=0
b=-3
we need value for a so as to solve the riddle
1) a^3=8
a=2 Sufficent
2) a^2=4
a=+-2 Two distinct values for a hence its Insufficient.
Answer A
a^(3b-b+6)=1
2b+6=0
b=-3
we need value for a so as to solve the riddle
1) a^3=8
a=2 Sufficent
2) a^2=4
a=+-2 Two distinct values for a hence its Insufficient.
Answer A
