[Guest]

Determine all possible pair(s) (x, y) of nonzero real numbers that satisfy this system of equations:

(3*x)^{log 3} = (7*y)^{log 7}, and:

7^{log x} = 3^{log y}

Note: All logarithms are considered in base ten.

Edited March 18, 2010 by K Sengupta

[Guest]

I see only one solution x=1/3 and y=1/7. Taking the log of the first equation. and the log of the second equation, we can eliminate log x and end up with log 7 + log y = 0

[Guest]

I found that if y = 3, x is approximately 16.33;

[Guest]

Taking the second equation we found that logx=logy*log3/log7

From the first equtation log3(log3+logx)=log7(log7+logy)

replacing logx=logy*log3/log7

We find that y=1/7

and x=1/3

Edited March 21, 2010 by Glosom

[dark_magician_92]

Determine all possible pair(s) (x, y) of nonzero real numbers that satisfy this system of equations:

(3*x)^{log 3} = (7*y)^{log 7}, and:

7^{log x} = 3^{log y}

Note: All logarithms are considered in base ten.

by eliminating log x or log y, we can solve the 2 eqns to find x=1/3, y=1/7