If x =-1 for the derivative of ln(x)=1/x does that imply that ln(-1) = -1? I wanted to find the solution of x for y=ln(x)-1/x=0 and my first intuition was to rearrange the equation to ln(x)=1/x and find the first derivative of the equation which gave me 1/x=-x^-2. I rearranged this equation to get x and it got me to x=-1. this made me ask if x=-1, does that imply that ln(-1)=-1? I understand that there are other ways to find x which is true for the equation. but I don't understand why this doesn't work and why ln(-1) is not equal to -1 even when x=-1. Does anybody have any ideas as to why this is?
 A: Here's where your argument is not valid.
When we solve an equation of the forme
$$f(x)=g(x)\text{ at } \Bbb R$$
this doesn't mean that the equality above is satisfied for all reals $ x $.
So, as you did, we cannot differentiate to get
$$f'(x)=g'(x)$$
As a counterexample, the equation
$$x^2=4$$
becomes
$$2x=0$$
which is an other equation.
A: The solution to $\ln x =\frac1x$ can be written as $$x\ln x =\ln(x^x)=1,$$
or $x^x=e.$ There must be a solution $1<x<2,$ since $x^x$ is increasing for $x>1$ and $1^1<e<2^2.$
There are no other real solutions, because when $0<x\leq 1,$ $x^x\leq 1,$ and $\ln x$ is not defined for $x\leq 0.$
If $y=\ln x$ then the equation becomes $x\ln x =ye^y=1,$ which means we can write the answer as $y=W(1)$ where $W$ is the Lambert W-function. Then $x=e^y=e^{W(1)}.$
It is unclear why taking the derivative helps you at all solve this question. In general, $f’(x_0)=g’(x_0)$ doesn’t help you solve $f(x)=g(x).$
Derivatives of a function are only defined where the function is defined. Anyway, the “anti-derivative” of $\frac1x$ is $\ln|x|,$ not $\ln x,$ when $x<0.$
