Showing the exponential and logarithmic functions are unique in satisfying their properties

The question asks to prove that there exists a unique function defined on $\Bbb R$ and satisfying the following conditions:

1) $f(1) = a$ $(a>0, a \neq 0)$

2) $f(x_1) \cdot f(x_2) = f(x_1 + x_2)$

3) $f(x) \rightarrow f(x_0)$ as $x \rightarrow x_0 $

Since the text goes into detail constructing the exponential function and proving these properties, I assume I only have to show uniqueness. i.e. showing that if functions $f$ and $g$ satisfy these properties, then $f=g$.

$f(1) = g(1)$ by property (1).

Then for $n\in \Bbb N$, $(f-g)(n)$ $=$$f(n) – g(n)$ $= f(1)^n -g(1)^n $ by property (2) and induction.

But $f(1) = g(1) = a > 0$ , which implies $f(1)^n -g(1)^n = 0$ and hence $f(n) =g(n)$

We have $f(1)-g(1)$ $=$ $f(\frac1n \cdot n) -g(\frac1n \cdot n)$ $=$$f(\frac1n)^n-g(\frac1n)^n = 0$.

Now $f(x) = f(\frac x2)^2 \ge 0$, so $f(\frac1n) =g(\frac1n)$.

$f(\frac mn ) -g (\frac mn ) $ $=$ $f(\frac1n)^m – g(\frac1n)^m$ which implies that $f(\frac mn) – g(\frac mn)$ $=0$.

For $x \in \Bbb R$, $\lim \limits_{\Bbb Q \in r \to x}$ $(f-g)(r) =0$: for any $\epsilon > 0$, we can choose a $\delta > 0$ such that $|r – x| < \delta $ and $|f(r) – g(r)| = 0 < \epsilon $.

By property 3 we also have $\lim \limits_{\Bbb Q \in r \to x}(f-g)(r) =\lim \limits_{\Bbb Q \in r \to x} f(r) – \lim \limits_{\Bbb Q \in r \to x} g(r) = f(x) – g(x)$.

Thus $f(x) = g(x)$ $\forall x \in \Bbb R$.

Is this approach correct?

A similar question is asked for the logarithmic function.

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