Question

Question #7e866

Answer 1

`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))`