# Logarithmic and exponential functions

This is an exercise from the book "Logaritmos" by Elon Lages Lima.

A bijection $$E:\mathbb{R} \rightarrow \mathbb{R}^+$$ is called an exponential funtion when its inverse $$F:\mathbb{R}^+ \rightarrow \mathbb{R}$$ is a logarithmic function. Prove that the bijetion $$E:\mathbb{R} \rightarrow \mathbb{R}^+$$ is an exponential function if, and only if:

a) $$E$$ is increasing, i.e., $$x < y \Rightarrow E(x) < E(y)$$;

b) $$E(x+y) = E(x)\cdot E(y)$$.

Additional info: in this book, the logarithmic function is defined as a function $$L: \mathbb{R}^+ \rightarrow \mathbb{R}$$ with the following properties:

i) $$x< y \Rightarrow L(x) < L(y)$$;

ii) $$L(xy) = L(x) + L(y), \forall x, y \in \mathbb{R}^+$$

I managed to prove the first implication, i.e., that $$E = F^{-1} \Rightarrow E$$ has properties a) and b).

I've been stuck with the second implication.

The inverse of any strictly increasing function exists and is strictly increasing; this general fact shows that (a) implies (i).

If (b) holds, and $$F=E^{-1}$$, then applying (b) with $$F(x)$$ and $$F(y)$$ in place of $$x$$ and $$y$$ yields $$E(F(x)+F(y)) = E(F(x))\cdot E(F(y)) = x\cdot y.$$ Now taking $$F$$ of both sides gives $$F(x) + F(y) = F(E(F(x)+F(y))) = F(x\cdot y)$$ which is (ii).

• We should show that if a) and b) hold, then $E = F^{-1}$. We can't assume that E and F are inverse. Commented Jun 15, 2019 at 20:17
• I think my understanding of the logic is correct; think about it again. Commented Jun 16, 2019 at 1:15
• I have some difficulties in understanding why you have assumed that $E = F^{-1}$. In my understanding, it should be shown that $(\Rightarrow)$ $E = F^{-1}$ implies that a) and b) hold, and $(\Leftarrow)$ if a) and b) hold, then $E = F^{-1}$. Could you you clarify you assumption that $E = F^{-1}$ Commented Jun 16, 2019 at 13:35
• @KirkLand It is given for both the $\Rightarrow$ and $\Leftarrow$ implications that $E$ is a bijection. So $E$ has an inverse. This answer gives that inverse the name $F.$ The answer does not assume (as you apparently did) that merely naming a function $F$ gives it any special properties (such as being a logarithm). Commented Jun 16, 2019 at 14:14
• @DavidK, thanks. Just checking if I understood: it's known that $E$ has an inverse. We show that $E^{-1}$ has properties (i) and (ii). Thus, $E^{-1}$ must be a logarithmic function. Hence, $E$ is exponential. Is that the core of the reasoning? Commented Jun 16, 2019 at 15:56

$$(\Rightarrow)$$ We assume that $$E$$ is exponential, i.e., $$E = F^{-1}$$. We must show that a) and b) hold.

Let's show a), i.e., $$E$$ is increasing. Take $$x_1, x_2, \in \mathbb{R}$$ with $$x_1 < x_2$$. Assume that $$E(x_1) \geq E(x_2)$$.

Since $$F$$ is increasing and $$E(x_1), E(x_2) \in \mathbb{R}^+$$, $$F(E(x_1)) \geq F(E(x_2))$$. Since $$E= F^{-1}$$ by assumption, it follows that $$x_1 \geq x_2$$, which is a contradiction. Therefore, $$E(x_1) < E(x_2)$$, i.e., a) holds.

Now let's prove that b) hold.

Let $$x, y \in \mathbb{R} = Im (F)$$. Thus, there exists $$x_0$$ and $$y_0 \in \mathbb{R}^+$$ such that $$F(x_0) = x$$ and $$F(y_0)= y$$ (by surjectivity of $$F$$).

Adding the two equations we have $$F(x_0) +F(y_0) = x + y$$. Since $$F$$ is logarithmic, $$F(x_0) +F(y_0) = F(x_0 \cdot y_0) = x + y$$

Applying $$E$$, it follows that $$E(x+y) = E(F(x_0 \cdot y_0)) = x_0 \cdot y_0$$.

However, $$F(x_0) = x \Leftrightarrow E(x) = x_0$$ and $$F(y_0) = y \Leftrightarrow E(y) = y_0$$.

Thus, $$E(x+y)= x_0 \cdot y_0 \Rightarrow E(x + y) = E(x) \cdot E(y)$$.

Therefore , b) holds.