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If 6^A = 2, 6^B = 5, and 6^Q = 15, then express Q in terms o

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If 6^A = 2, 6^B = 5, and 6^Q = 15, then express Q in terms o

by Mike@Magoosh » Wed Nov 05, 2014 10:21 am
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 OA & OE of this particular problem, as well as a full set of challenging problems on exponents, see:
https://magoosh.com/gmat/2014/challengi ... and-roots/

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by GMATinsight » Wed Nov 05, 2014 10:54 am
Mike@Magoosh 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 OA & OE of this particular problem, as well as a full set of challenging problems on exponents, see:
https://magoosh.com/gmat/2014/challengi ... and-roots/

Mike :-)
Answer: Option E
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by Mathsbuddy » Mon Nov 10, 2014 11:41 am
As GMat problems require a straight forward answer, you can solve this quickly as follows:

The given integers are 2, 5, 6 and 15.
Now, by inspection, find a common multiplicative relationship between them:

5/2 = 15/6 (= 2.5)

So, 6^B/6^A = 6^Q/6^1
Index laws give us:
6^(B-A) = 6^(Q-1)
By equivalence of exponents:
B-A = Q-1
Therefore Q = B - A + 1 (JOB DONE!)
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by Mathsbuddy » Mon Nov 10, 2014 11:44 am
Even quicker:

Using 15 = 6 x 5/2
gives
6^Q = 6^1 x 6^B/6^A
Index laws give us:
6^Q = 6^(1 + B - A)
Therefore Q = B - A + 1
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by Matt@VeritasPrep » Mon Nov 10, 2014 11:59 am
How about an easier way with less jargon? (Note: for whatever reason 'q' is not in the ASCII superscript alphabet, so I'm using 6á´½ to mean 6^q.)

6á´½ = 15 = 30/2 = (6*5)/2

Since we know 2 and 5 in terms of powers of 6, let's replace 2 and 5 with those powers.

6á´½ = (6*6á´®)/6á´¬
6ᴽ = (6ᴮ�¹)/6ᴬ
6ᴽ = 6ᴮ�¹�ᴬ

So Q = B + 1 - A, or B - A + 1.

This is neat, but I don't know if the GMAT would ask a question that's so trivial for students who already know their logarithms and so challenging for students who don't. (The GMAT tends to avoid questions that are easily solved with methods from trig and calculus, but hey, you never know!)
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by Mathsbuddy » Tue Nov 11, 2014 4:42 am
Alternatively for those who like logarithms:

Using log x = log x (base 6):

B = log 5
A = log 2
1 = log 6

Q = log 15 = log(5/2 * 6)
Using logarithm laws, this becomes:
Q = log5 - log 2 + log 6 = B - A + 1
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