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 establish that the function $f(x) = \log_a x$ is a bijection, it must be demonstrated as both injective and surjective. 1. Injectivity: A function is injective if $f(x_1) = f(x_2)$ implies $x_1 = x_2$. Assuming $f(x_1) = f(x_2)$, we have: \[ \log_a x_1 = \log_a x_2. \] Using logarithm properties, this simplifies to: \[ x_1 = x_2. \] Thus, $f(x) = \log_a x$ is injective. 2. Surjectivity: A function is surjective if for every $y \in \mathbb{R}$, there exists an $x \in \mathbb{R}^+$ such that $f(x) = y$. Let $y$ be any real number. We seek an $x \in \mathbb{R}^+$ satisfying: \[ \log_a x = y. \] This equation is equivalent to: \[ x = a^y. \] Since $a>0$ and $a eq 1$, $a^y$ is always a positive real number for any $y \in \mathbb{R}$. Therefore, $f(x) = \log_a x$ is surjective. As $f(x)$ is both injective and surjective, it is bijective. %Quuciktip