Question

If $f : \mathbb{R}^+ \to \mathbb{R}$ is defined as $f(x) = \log_a x$ where $a>0$ and $a \neq 1$, prove that $f$ is a bijection. (R$^+$ is the set of all positive real numbers.)

Hint:

A function is a bijection if it is both injective (one-to-one) and surjective (onto).

Answer 1

To prove that the function $f(x) = \log_a x$ is a bijection, we need to show that it is both injective (one-to-one) and surjective (onto). 1. Injectivity: - For a function to be injective, if $f(x_1) = f(x_2)$, then $x_1 = x_2$ must hold. - Let $f(x_1) = f(x_2)$, i.e., \[ \log_a x_1 = \log_a x_2. \] - By the property of logarithms, we can rewrite this as \[ x_1 = x_2. \] Therefore, the function $f(x) = \log_a x$ is injective. 2. Surjectivity: - A function is surjective if for every $y \in \mathbb{R}$, there exists $x \in \mathbb{R}^+$ such that $f(x) = y$. - Let $y \in \mathbb{R}$ be arbitrary. - We need to find $x \in \mathbb{R}^+$ such that \[ \log_a x = y. \] - This is equivalent to \[ x = a^y, \] which is always a positive real number for any $y \in \mathbb{R}$ (since $a>0$ and $a \neq 1$). Therefore, $f(x) = \log_a x$ is surjective. Since the function $f(x)$ is both injective and surjective, it is bijective. %Quuciktip