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

1 Answer
Nov 10, 2015

Find out the turning points of #f(x) = x^2-x-ln(x)# and hence determine that the only solution of #f(x) = 0# and hence the only solution of #x^2-x = ln(x)# is #x = 1#

Explanation:

Let #f(x) = x^2-x-ln(x)#

Then #f'(x) = 2x-1-1/x#

Then:

#xf'(x) = 2x^2-x-1 = (x-1)(2x+1)#

Hence #f'(x) = 0# when #x = -1/2# or #x = 1#

As a Real valued function, #ln(x)# is only defined for #x > 0#, so the only turning point of #f(x)# is where #x = 1#.

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

So the only Real zero of #f(x)# is where #x = 1#
graph{x^2-x-ln(x) [-9.08, 10.92, -1.68, 8.32]}