Question

Consider the logarithm function \(\ln :(0, \infty) \rightarrow \mathbb{R}\). Decide whether this function is injective and whether it is surjective.

Short answer

The natural logarithm function \(\ln :(0, \infty) \rightarrow \mathbb{R}\) is both injective and surjective.

Step-by-step solution

Checking injectivity

Observe two arbitrary elements \(x_1\) and \(x_2\) from (0, \infty). Suppose \(\ln (x_1) = \ln (x_2)\). By properties of logarithm, it must then be that \(x_1=x_2\). Hence, the function \(\ln\), is injective since every element of (0, \infty) maps to a unique element in \(\mathbb{R}\).

Checking surjectivity

Consider any \(y\) in \(\mathbb{R}\). Clearly, there exists \(x\) in (0, \infty) such that \(y = \ln(x)\) (choose \(x = e^y\)). Hence, the function \(\ln\) is surjective since every element of \(\mathbb{R}\) is the image of some element in (0, \infty).