Properties of Natural Logarithm

Theorem
Let $$x \in \mathbb{R}$$ be a real number such that $$x > 0$$.

Let $$\ln x$$ be the natural logarithm of $$x$$.

Then:
 * $$\ln 1 = 0$$;
 * $$D \ln x = \frac 1 x$$
 * The function $$f \left({x}\right) = \ln x: x > 0$$ is strictly increasing and concave;
 * $$\ln x \to +\infty$$ as $$x \to +\infty$$;
 * $$\ln x \to -\infty$$ as $$x \to 0^+$$.

Proof

 * $$\ln 1 = 0$$:

From the definition, $$\ln x = \int_1^x \frac {dt} t$$.

From Integral on Zero Interval, $$\ln 1 = \int_1^1 \frac {dt} t = 0$$.


 * $$D \ln x = \frac 1 x$$

From the definition, $$\ln x = \int_1^x \frac {dt} t$$.

Thus from the Fundamental Theorem of Calculus we have that $$\ln x$$ is a primitive of $$\frac 1 x$$ and hence $$D \ln x = \frac 1 x$$.


 * $$\ln x$$ is strictly increasing and concave:

From the above $$D \ln x = \frac 1 x$$, which is strictly positive on $$x > 0$$.

From Derivative of Monotone Function it follows that $$\ln x$$ is strictly increasing on $$x > 0$$.

From the Power Rule for Derivatives: Integer Index, $$D^2 \ln x = D \frac 1 x = \frac {-1} {x^2}$$.

Thus $$D^2 \ln x$$ is strictly negative on $$x > 0$$ (in fact is strictly negative for all $$x \ne 0$$).

Thus from Derivative of Monotone Function, $$D \frac 1 x$$ is strictly decreasing on $$x > 0$$.

So from Derivative of Convex or Concave Function, $$\ln x$$ is concave on $$x > 0$$.


 * $$\ln x \to +\infty$$ as $$x \to +\infty$$:

This follows directly from Integral of Reciprocal is Divergent.


 * $$\ln x \to -\infty$$ as $$x \to 0^+$$.

This follows directly from Integral of Reciprocal is Divergent.