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 
Senior Member Joined: Apr 2013 Posts: 425 Thanks: 24  If it is a system of equations, then apply the "" of the two equations and you'll get a system of equations with the unknowns and .
Last edited by skipjack; April 6th, 2014 at 05:39 AM. 
April 6th, 2014, 05:47 AM  #4 
Global Moderator Joined: Dec 2006 Posts: 20,835 Thanks: 2162 
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  
Math Team Joined: Dec 2006 From: Lexington, MA Posts: 3,267 Thanks: 408  Hello, mared! I'll get you started . . . Quote:
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]: $\displaystyle \;\;\;\begin{array}{cc}\log\left(x^{\log5}\right) \;=\;\log\left(y^{\log3}\right) \\ \log5\log x \;=\;\log3\log y \\ \log5\log x  \log3\log y \;=\;0 & [4] \end{array}$ $\displaystyle \text{Let: }\:\begin{Bmatrix}a \,=\,\log3 \\ b \,=\,\log5 \end{Bmatrix}\quad \begin{Bmatrix}X \,=\,\log x \\ Y \,=\,\log y \end{Bmatrix}$ $\displaystyle \text{Then [3] and [4] become: }\:\begin{Bmatrix}aX  bY &=& b^2a^2 \\ bX  aY &=& 0 \end{Bmatrix}$ Solve the system and backsubstitute.  
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).
