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 October 25th, 2013, 04:49 PM #1 Newbie   Joined: Oct 2013 Posts: 3 Thanks: 0 Logarithms question p, q > 0 (log b_p q) + (log b_q p) = 5/2 pq = 54sqrt2 What is (p+q) /3? Can anyone give me pointers on where to start?
 October 25th, 2013, 05:57 PM #2 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,968 Thanks: 1152 Math Focus: Elementary mathematics and beyond Re: Logarithms question The notation is confusing. What is the base of each logarithm and what is the argument of each logarithm?
October 25th, 2013, 06:21 PM   #3
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Re: Logarithms question

Hello, cube198!

I think I understand what you are saying . . .

Quote:
 $\begin{array}{ccc}p,\,q\: >\: 0 && [1]\\ \\ \\ \log_p q\,+\,\log_q p \:=\: \frac{5}{2} && [2] \\ \\ \\ pq \:=\: 54\sqrt{2} && [3] \end{array}$ $\text{Find }\,\frac{p\,+\,q}{3}$

$\text{From [2], we have: }\:\log_pq \,+\,\frac{1}{\log_pq} \:=\:\frac{5}{2}$

$\text{Multiply by }2\,\!\log_pq: \;\;2(\log_pq)^2\,+\,2 \:=\:5\,\!\log_pq$

[color=beige]. . . . . . . . . . .[/color]$2(\log_pq)^2\,-\,5\,\!\log_pq\.+\.2 \:=\:0$

$\text{Factor: }\;\;\;\;\;\; (2\,\!\log_pq\,-\,1)(\log_pq\,-\,2) \:=\:0$

$\text{W\!e have: }$

$\begin{Bmatrix} 2\,\!\log_pq \,-\,1 \:=\:0 && \Rightarrow && \log_pq \:=\:\frac{1}{2} && \Rightarrow && p^{\frac{1}{2}} \:=\:q \\ \\ \\
\log_pq\,-\,2 \:=\:0 && \Rightarrow && \log_pq \:=\:2 && \Rightarrow && p^2 \:=\:q
\end{Bmatrix}$

$\text{Substitute }q \,=\,p^{\frac{1}{2}}\text{ into [3]: }\;p\left(p^{\frac{1}{2}}\right) \,=\,54\sqrt{2} \;\;\;\Rightarrow\;\;\;p^{\frac{3}{2}} \:=\:\left(3\sqrt{2}\right)^3$

$\text{Hence: }\:p \,=\,(3\sqrt{2})^2 \;\;\;\Rightarrow\;\;\; \fbox{p \:=\:18 \;\;\;\Rightarrow\;\;\;q \,=\,3\sqrt{2}}$

$\text{Substitute }q \,=\,p^2\text{ into [3]: }\:p(p^2) \,=\,54\sqrt{2} \;\;\;\Rightarrow\;\;\;p^3 \:=\:(3\sqrt{2})^3$

$\text{Hence: }\:\fbox{p \,=\,3\sqrt{2} \;\;\;\Rightarrow\;\;\;q \,=\,18}$

 October 26th, 2013, 02:27 AM #4 Global Moderator   Joined: Dec 2006 Posts: 21,036 Thanks: 2273 Either way, (p + q)/3 = 6 + ?2.
 October 26th, 2013, 07:45 AM #5 Newbie   Joined: Oct 2013 Posts: 3 Thanks: 0 Re: Logarithms question Thank you very much to everyone that helped!

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