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If 6^A = 2, 6^B = 5, and 6^Q = 15, then express Q in terms of A and/or [#permalink]
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05 Nov 2014, 10:17
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If \(6^A = 2\), \(6^B = 5\), and \(6^Q = 15\), then express Q in terms of A and/or B. (A) 3B (B) B + 3 (C) 5A + B (D) \(B^2  2B\) (E) B  A + 1For the solution to this, as well as for a set of challenging problems on exponents, see: https://magoosh.com/gmat/2014/challengi ... androots/Mike
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Re: If 6^A = 2, 6^B = 5, and 6^Q = 15, then express Q in terms of A and/or [#permalink]
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06 Nov 2014, 00:29
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mikemcgarry wrote: If \(6^A = 2\), \(6^B = 5\), and \(6^Q = 15\), then express Q in terms of A and/or B. (A) 3B (B) B + 3 (C) 5A + B (D) \(B^2  2B\) (E) B  A + 1For the solution to this, as well as for a set of challenging problems on exponents, see: https://magoosh.com/gmat/2014/challengi ... androots/Mike 6^A=2 1) 6^B=5 2) 6^Q= 15 3) multiplying 1,2,3 together, we have 6^(A+B+Q)= 2*5*15 = 2*3*5^2 =6*5^2 from 2, we have 5^2= 6^2B =6*6^2B =6^2B+1 thus A+B+Q=2B+1 Q=BA+1



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Re: If 6^A = 2, 6^B = 5, and 6^Q = 15, then express Q in terms of A and/or [#permalink]
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11 Nov 2014, 01:43
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\(6^q = 15 = \frac{15 * 2}{2} = \frac{30}{2} = \frac{5 * 6}{2}\) .................. (1) \(6^A = 2, 6^B = 5\)>> Placing these values in (1) \(6^q = \frac{6^b * 6}{6^a}\) \(6^q = 6^{(b+1a)}\) Bases are same; equating the powers q = b+1a Answer = E
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Re: If 6^A = 2, 6^B = 5, and 6^Q = 15, then express Q in terms of A and/or [#permalink]
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06 Nov 2017, 10:06
mikemcgarry wrote: If \(6^A = 2\), \(6^B = 5\), and \(6^Q = 15\), then express Q in terms of A and/or B. (A) 3B (B) B + 3 (C) 5A + B (D) \(B^2  2B\) (E) B  A + 1For the solution to this, as well as for a set of challenging problems on exponents, see: https://magoosh.com/gmat/2014/challengi ... androots/Mike Hi see if all can be brought to base 6..now we have 2,5,15 and these can be arranged as \(\frac{15}{5}*2=6\) substitute corresponding values... \(\frac{15}{5}*2=\frac{6^Q}{6^B}*6^A=6^{Q+AB}=6^1........Q+AB=1........Q=B+1A\) E
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Re: If 6^A = 2, 6^B = 5, and 6^Q = 15, then express Q in terms of A and/or [#permalink]
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06 Nov 2017, 11:07
6^A=2, 6^B=5, and 6^Q=15 applying log base 6 (log6) to both sides of the equations log6(6^A) = log6(2) log6(6^B)=log6(5) log6(6^Q)=log6(15) using log properties
A = log6(2) B = log6(5) Q = log6(15) and just scrolling the answers down for anyone who knows logs first answer to check is E (as others lead to much greater values of Q) log6(15) = log6(5)  log6(2) + 1 1 is log6(6) so we can rewrite log6(15) = log6(5)  log6(2) + log6(6) > applying another log property log6(15) = log6 ((5/2)*6) = log6 (15)




Re: If 6^A = 2, 6^B = 5, and 6^Q = 15, then express Q in terms of A and/or
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06 Nov 2017, 11:07






