Question

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

Answer 1

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]}