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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)