# solve for x,yx+log base 2 y=3; x^2 + (log base 2 y)^2=5

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You need to solve for x and y simultaneous equations.

Notice that `x^2 + (log_2 y)^2 = (x + log_2 y)^2 - 2x*log_2 y`

Substituting 5 for `x^2 + (log_2 y)^2 ` and 3 for`x + log_2 y` yields:

`5 = 9 - 2x*log_2 y =gt 5 - 9 = - 2x*log_2 y =gt -4 = - 2x*log_2 y`

`x*log_2 y = 2`

You notice that the problem provides the sum and the product of two numbers, hence you should come up with substitution: `x +` `log_2 y =q` and `x*log_2 y = ` `p` such that:

`w^2 - qw+ p = 0`

You should use the quadratic formula such that:

`w_(1,2) = (q+-sqrt(q^2 - 4p))/2`

Substituting 3 for q and 2 for p yields:

`w_(1,2) = (3+-sqrt(9 - 8))/2 =gt w_(1,2) = (3+-1)/2`

`w_1 = 2; w_2 = 1`

**Hence, the solutions to equations are either `x = 2` and `log_2 y = 1` => `y = 2^1=2, ` or `x = 1` and`log_2 y = 2 =gt y = 2^2 = 4.` **