# Solve for x, yx^2+y^2=5 log base 2 x -log base 4 y = 1

sciencesolve | Teacher | (Level 3) Educator Emeritus

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You need to find the solution to the simultaneous equations, hence you should start with the second equation by bringing the logarithms to a common base such that:

`log_4y = (log_2 y)/(log_2 4) =gt log_4 y = (log_2 y)/(log_2 (2^2))`

You should use the properties of logarithms such that: `(log_2 (2^2)) = 2(log_2 2)`

You need to substitute 1 for `(log_2 2)`  hence:

`(log_2 (2^2)) = 2`

`log_4 y = (log_2 y)/2`

You need to substitute `(log_2 y)/2`  for `log_4 y`  in `log_2 x + log_4 y = 1`  such that:

`log_2 x + (log_2 y)/2 = 1`

You need to bring the terms to a common denominator such that:

`2log_2 x + (log_2 y) = 2 `

`log_2 (x^2) + (log_2 y) = 2`

Notice that logarithms have like bases, hence you should convert the sum of logarithms in the logarithm of product such that:

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

`x^2 = 4/y`

You should substitute `4/y`  for `x^2`  in the first equation `x^2+y^2=5`  such that:

`4/y+y^2 = 5 `

You need to bring the terms to a common denominator such that:

`4+ y^3 - 5y = 0`

`y^3 - 5y + 4 = 0`

Notice that substituting 1 for y yields:

`1-5+4 = 0`

`y^3 - 5y + 4 = (y-1)(ay^2+by+c)`

`y^3 - 5y + 4 = ay^3 + y^2(b-a) + y(c-b)- c`

Equating the coefficients of like powers yields:

`a=1`

`b-a=0 =gt b-1=0 =gt b=1`

`c-b=-5 =gt c= -4`

`ay^2+by+c = y^2 + y - 4`

Hence, you need to solve the equation `y^2 + y - 4 = 0`  to find the next two zeroes such that:

`y_(2,3) = (-1+-sqrt(1+16))/2`

`y_(2,3) = (-1+-sqrt17)/2`

You need to substitute 1 for y in `x^2 = 4/y`  to find x such that:

`x_(1,2) = +-2`

You need to substitute `(-1+sqrt17)/2`  for y in `x^2 = 4/y`  to find x such that:

`x_(3,4) = +-2sqrt2/sqrt(-1+sqrt17)`

Since, you need to keep only the positive values, the solutions to system of equations are `(2,1) ` and `((2sqrt2/sqrt(-1+sqrt17),(-1+sqrt17)/2).`