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April 5th, 2014, 11:11 PM   #1
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(3x)^log3 =(5y)^log5
x^log5 = y^log3
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April 5th, 2014, 11:22 PM   #2
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April 5th, 2014, 11:40 PM   #3
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Quote:
Originally Posted by mared View Post
(3x)^log3 =(5y)^log5
x^log5 = y^log3
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.
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April 6th, 2014, 05:47 AM   #4
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The first equation is satisfied by (x, y) = (1/3, 1/5) or (x, y) = (0, 0).
What about the second?
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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.

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April 6th, 2014, 05:55 PM   #6
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The second equation is satisfied by (x, y) = (1, 1) or (x, y) = (0, 0).
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