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
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Mike McGarry
Magoosh Test Prep


Education is not the filling of a pail, but the lighting of a fire. — William Butler Yeats (1865 – 1939)

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

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)

Used the options to get to the answer:

1) => \(6^(3B)\)
=> \((6^B)*(6^B)*(6^B)\)
=> 5*5*5=125
So incorrect

2) \((6^(B+3))\)
=> \((6^B)*(6^3)\)
=> 5*216=1080
So incorrect

3) \((6^(5A+B))\)
=> \((6^5A)*(6^B)\)
=> \((2^5)*5\)
=> 32*5
=> 160
So incorrect

4) (6^\(B^2 - 2B\))
=> \(\frac{(6^(B^2))}{(6^2B)}\)
=> \(\frac{(5^B)}{25}\)
Cant be calculated with the given information

5) 6^(B-A+1)
=> \(\frac{((6^B)*6)}{(6^A)}\)
=> \(\frac{(5*6)}{2}\)
=> 15
So correct and the answer is E
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