Question #7e866

1 Answer
namralos
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))#

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