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# If 6^A = 2, 6^B = 5, and 6^Q = 15, then express Q in terms of A and/or

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Magoosh GMAT Instructor
Joined: 28 Dec 2011
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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, 11:17
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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 + 1

For the solution to this, as well as for a set of challenging problems on exponents, see:
https://magoosh.com/gmat/2014/challengi ... and-roots/

Mike
[Reveal] Spoiler: OA

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Mike McGarry
Magoosh Test Prep

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Kudos [?]: 8790 [1], given: 106

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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, 01:29
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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 + 1

For the solution to this, as well as for a set of challenging problems on exponents, see:
https://magoosh.com/gmat/2014/challengi ... and-roots/

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=B-A+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, 02:43
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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)

$$6^q = \frac{6^b * 6}{6^a}$$

$$6^q = 6^{(b+1-a)}$$

Bases are same; equating the powers

q = b+1-a

Answer = E
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Kudos [?]: 2723 [4], given: 193

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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, 11:06
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 + 1

For the solution to this, as well as for a set of challenging problems on exponents, see:
https://magoosh.com/gmat/2014/challengi ... and-roots/

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...
$$\frac{15}{5}*2=\frac{6^Q}{6^B}*6^A=6^{Q+A-B}=6^1........Q+A-B=1........Q=B+1-A$$
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, 12:07
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)

Kudos [?]: 14 [0], given: 126

Re: If 6^A = 2, 6^B = 5, and 6^Q = 15, then express Q in terms of A and/or   [#permalink] 06 Nov 2017, 12:07
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# If 6^A = 2, 6^B = 5, and 6^Q = 15, then express Q in terms of A and/or

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