# How do you solve #x^2-x= lnx#?

##### 1 Answer

Nov 10, 2015

Find out the turning points of

#### Explanation:

Let

Then

Then:

Hence

As a Real valued function,

#f(1) = 1^2-1-ln(1) = 1 - 1 - 0 = 0#

So the only Real zero of

graph{x^2-x-ln(x) [-9.08, 10.92, -1.68, 8.32]}