How to prove that $f(x) = a^x$ is a continuous bijection? I need to prove that $f(x) = a^x $ is a continuous bijection from $\mathbb{R}$ to $\mathbb{R}^*$ for $a>0$, $a\neq 1$. For injectivity, I did:
$$a^x = a^y \rightarrow \log a^x = \log a^y \rightarrow x\log a = y\log a \rightarrow x = y$$
(i can divide both sides by $\log a$ because $a\neq 1\implies \log a \neq 0$.
For surjectivity, I need to prove that there is always a solution for $a^x = y$. If I apply log to both sides: $x\log a = \log y\implies x = \frac{\log y}{\log a} = \log_a y$. Is this it? 
For continuity, can I just use continuity of $\log$ since $a^x = e^{\log a ^x}$?
 A: You have to prove that $f(x) = a^{x}, a > 0, a \neq 1$ is a bijection from $\mathbb{R}$ to $\mathbb{R}^{+}$. It is best to prove first that $f(x)$ is continuous for all $x$. In what follows we keep $a > 1$ and the case for $0 < a < 1$ is handled by noting that $f(x) = 1/(1/a)^{x}$ and $1/a > 1$.
A typical definition for $a^{x}$ is $f(x) = a^{x} = \exp(x\log a)$ and hence continuity of $f(x)$ follows from the continuity of $\exp(x)$. Next note that $$a^{x} = \exp(x\log a) \geq 1 + x\log a$$ and by taking limit as $x \to \infty$ we get $a^{x} \to \infty$ as $x \to \infty$. Noting that $f(x) = 1/a^{-x}$ we see that $f(x) \to 0$ as $x \to -\infty$. By intermediate value theorem for continuous functions $f$ takes all values in $\mathbb{R}^{+}$. Also note that $f(x)/f(y) = a^{x - y} \neq 1$ unless $x = y$. Hence $f$ is one-one. It follows that $f$ is a bijection.
A: Here's a hint for another way of proving this. Another way to show a function is a bijection is to show it has an inverse function. Here though you'll want to be careful with your domains and ranges/codomains. I'm not familiar with the notation $\mathbb{R}^*$ but I assume this just means all positive real numbers. The inverse function is thus going to be a function that has this as its domain. You basically did this procedure of finding the inverse when you found that $a^x=y$ leads to $x=log_a(y)$. 
