**I think the question was supposed to be: x^log 2x +x^log x= 2**

Now in x^log 2x +x^log x= 2 write (x^ log x) as y

=> y^2 + y = 2

=> y^2 + y - 2 = 0

=> y^2 + 2y - y - 2 = 0

...

## Unlock

This Answer NowStart your **48-hour free trial** to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

Already a member? Log in here.

**I think the question was supposed to be: x^log 2x +x^log x= 2**

Now in x^log 2x +x^log x= 2 write (x^ log x) as y

=> y^2 + y = 2

=> y^2 + y - 2 = 0

=> y^2 + 2y - y - 2 = 0

=> y( y + 2) - 1(y + 2) = 0

=> (y - 1)(y +2) = 0

So y = 1 and y = -2

x^ log x = 1 and x^ log x = -2

Now x^ log x = -2 is not possible as x would have to be negative and in that case log x would not be defined.

So x^log x = 1

=> log x = 0

=> x = 10^0

=> x = 1

**Therefore x = 1**