# Solve the following system of equations.`5^logx- 3^logy=0``(3x)^log3-(5y)^log5=0`

## Expert Answers

The following system of equations has to be solved:

`5^logx- 3^logy=0`...(1)

`(3x)^log3-(5y)^log5=0`...(2)

`5^log x = 3^log y`

take the log of both the sides

=> `log 5*log x = log 3*log y`

=> `log x = (log 3/log 5)*log y`

`(3x)^log3-(5y)^log5 = 0`

=> `(3x)^log3 = (5y)^log5`

take the log of both the sides

=> `log 3*log(3x) = log 5*log(5y)`

=> `log 3*(log 3 + log x) = log 5*(log 5 + log y)`

Substitute `log x = (log 3/log 5)*log y`

=> `log 3*(log 3 + (log 3/log 5)*log y) = log 5*(log 5 + log y)`

=> `(log 3)^2 + ((log 3)^2/log 5)*log y = (log 5)^2 + log 5*log y`

=> `log y((log 3)^2/log 5 - log 5) = (log 5)^2 - (log 3)^2`

=> `log y ~~ -0.6989`

`log x ~~ -0.4771`

`x = 1/3`

`y = 1/5`

The solution of the set of equations is `x = 1/3` and `y = 1/5`

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