Question #7e866

1 Answer
Jul 20, 2017

x^2 + 2xy = log(x^x) + 2log(x^y)
Bring the powers down
x^2 + 2xy = xlogx + 2ylogx
Separate the terms that have y in them from those that do not:
2xy - 2ylogx = xlogx - x^2
Factor everything on the left except for y:
y(2x - 2logx) = xlogx - x^2
Divide by the expression in parentheses:
y = (xlogx - x^2)/(2x - 2logx)

Now, the other expression simplifies slightly:

Let 2^(2/y) = 4^(1/y).

Since we have discovered a value for y, we may take its reciprocal, so that:

1/y = (2x - 2logx)/(xlogx - x^2)

Finally, exponentiate in base 4:

4^(1/y) = 4^((2x - 2logx)/(xlogx - x^2))