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 April 5th, 2014, 11:11 PM #1 Senior Member   Joined: Nov 2013 Posts: 434 Thanks: 8 log (3x)^log3 =(5y)^log5 x^log5 = y^log3
 April 5th, 2014, 11:22 PM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs What have you tried?
April 5th, 2014, 11:40 PM   #3
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Quote:
 Originally Posted by mared (3x)^log3 =(5y)^log5 x^log5 = y^log3
If it is a system of equations, then apply the "$\log$" of the two equations and you'll get a system of equations with the unknowns $\log(x)$ and $\log(y)$.

Last edited by skipjack; April 6th, 2014 at 05:39 AM.

 April 6th, 2014, 05:47 AM #4 Global Moderator   Joined: Dec 2006 Posts: 21,019 Thanks: 2253 The first equation is satisfied by (x, y) = (1/3, 1/5) or (x, y) = (0, 0). What about the second?
April 6th, 2014, 05:03 PM   #5
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Hello, mared!

I'll get you started . . .

Quote:
 $\displaystyle \begin{array}{cccc}(3x)^{\log3} &=& (5y)^{\log5} & [1] \\ x^{\log5} &=& y^{\log3} & [2]\end{array}$

Take the log of equation [1]:

$\displaystyle \;\;\;\begin{array}{cc}\log(3x)^{\log3} \:=\:\log(5y)^{\log5} \\ \log3\log(3x) \;=\;\log5\log(5y) \\ \log3(\log3 + \log x) \;=\;\log5(\log5+\log y) \\ (\log3)^2 + \log3\log x \;=\;(\log5)^2 + \log5\log y \\ \log3\log x - \log5\log y \;=\;(\log5)^2 - (\log3)^2 & [3]\end{array}$

Take the log of equation [2]:

$\displaystyle \;\;\;\begin{array}{cc}\log\left(x^{\log5}\right) \;=\;\log\left(y^{\log3}\right) \\ \log5\log x \;=\;\log3\log y \\ \log5\log x - \log3\log y \;=\;0 & [4] \end{array}$

$\displaystyle \text{Let: }\:\begin{Bmatrix}a \,=\,\log3 \\ b \,=\,\log5 \end{Bmatrix}\quad \begin{Bmatrix}X \,=\,\log x \\ Y \,=\,\log y \end{Bmatrix}$

$\displaystyle \text{Then [3] and [4] become: }\:\begin{Bmatrix}aX - bY &=& b^2-a^2 \\ bX - aY &=& 0 \end{Bmatrix}$

Solve the system and back-substitute.

 April 6th, 2014, 05:55 PM #6 Senior Member   Joined: Mar 2014 Posts: 112 Thanks: 8 The second equation is satisfied by (x, y) = (1, 1) or (x, y) = (0, 0).

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