# Solve simultaneos equations `x^2+y^2=5` `log_2(x)-log_4(y)=1`x y

sciencesolve | Teacher | (Level 3) Educator Emeritus

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You should perform several operations in the second equation, using logarithmic identities such that:

`log_2 x = 1 + log_4 y => log_2 x = log_4 4 + log_4 y`

`log_2 x = log_4 (4*y)`

You need to transform the base 4 in base 2 such that:

`log_4 (4*y) = (log_2 (4*y))/(log_2 4) => log_4 (4*y) = (log_2 (4*y))/2`

Hence, substituting `(log_2 (4*y))/2`  for `log_4 (4*y)`  in the second equation yields:

`log_2 x = (log_2 (4*y))/2 =>2log_2 x = (log_2 (4*y))`

Using the logarithmic identity `n*log x = log (x^n)`  yields:

`log_2 x^2 = log_2 (4*y) => x^2 = 4y`

Notice that the first equation provides the information that `x^2 + y^2 = 5` , hence, you need to substitute `4y`  for `x^2`  in the first equation such that:

`4y + y^2 = 5 => y^2 + 4y - 5 = 0`

`y_(1,2) = (-4+-sqrt(16 + 20))/2 => y_(1,2) = (-4+-sqrt36)/2`

`y_(1,2) = (-4+-6)/2 => y_1 = 1 ; y_2 = -5`

You should notice that y needs to have a positive value since it is the argument of logarithm, hence, you need to keep only the positive solution `y = 1` .

Substituting 1 for y in the first equation yields:

`x^2 + 1^2 = 5 => x^2 = 5 -1 => x^2 = 4 => x_(1,2) = +-2`

Notice that x needs to have a positive value since it is the argument of logarithm, hence you need to keep `x = 2` .

Hence, evaluating the solution to the system of equations yields `x =2`  and `y = 1` .

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