Prove logarithm rules using definition as the inverse exponential

Question

Problem $3$. Show that $\operatorname{exp} : \Bbb R \to (0, \infty)$ is bijective. Its inverse function is called the (natural) logarithm $\log : (0, \infty) \to \Bbb R$. Verify the logarithm rules: $$\log(ab) = \log(a) + \log(b), \quad \log(a^n) = n \log(a), n \in \Bbb Z, \quad \log(1) = 0$$ Also show that $\log$ is a continuous function.

I have already proved that $\exp$ is bijective on this domain and range, so I know the inverse $\log$, and I want to prove the logarithm rules that are listed using the definition of the logarithm as the inverse exponential function.

How can I do this?

Answer 1

Well since $a,b\in\mathbb{R}^+$ (positive reals), you know that $a= \exp(x)$ and $b=\exp(y)$ for some $x,y\in\mathbb{R}$. Then use the fact that $\log$ is the inverse function to $\exp$ in conjunction with properties of exponentials.

To see the second one, note that $n\log(a) = \underbrace{\log(a)+\cdots+\log(a)}_{\text{n times}}$.

I think the last one should be clear.

To see how $\log$ is continuous, we want that if $|x-y|<\delta$, then $|\log(x)-\log(y)|<\varepsilon$ (where $\delta$ depends on $\varepsilon$). Since $x,y>0$, we know that $x=\exp(x_0)$ for some $x_0\in\mathbb{R}$ and likewise $y = \exp(y_0)$. Then this recasts our if-then as:

If $|\exp(x_0)-\exp(y_0)|<\delta$, then $|x_0-y_0|<\varepsilon$.

However you know that $\exp$ is a bijection and so the following is true:

$|x_0-y_0|<\delta$ (for some appropriate choice of $\delta$) if and only if $|\exp(x_0)-\exp(y_0)|<\varepsilon$.

Do you see how this helps you?