Showing different definitions of exponentiation are equivalent

Question

Suppose we define $\exp(x)$ as the unique function $f:\mathbb{R} \rightarrow \mathbb{R_+}$ satisfying $f(0) = 1$ and $f'(x) = f(x)$ for all $x \in \mathbb{R}$. We then define its inverse $f:\mathbb{R_+} \rightarrow \mathbb{R}$ denoted $\log(x)$. Then, define $b^x$ for $b>0$ as $\exp(x\log b)$.

Then, how can we show that the "usual" properties of exponentiation are satisfied for integer and rational values of $x$ and $b$. Namely, $\prod_{i=1}^{n}{b} = b^n$, $b^{p/q} = \sqrt[q]{b^p}$ (where $\sqrt[q]{x}$ for $x>0$ has been appropriately defined as the unique positive number $m$ satisfying $\prod_{i=1}^{q}{m} = x$; of course we needn't consider negative arguments since $b>0$ ), $b^qb^p = b^{p+q}, $$(b^q)^p = b^{pq}$, etcetera.

In other words, how can we show this somewhat artificial definition is equivalent to the notion of exponentiation that we all are familiar with.

Answer 1

First let's show that $\exp(x+y)=\exp(x)\exp(y)$. For that, fix $y$ and consider $f(x)=\frac{\exp(x+y)}{\exp(y)}$. Then we easily see that $f(0)=1$ and $f'(x)=f(x)$, so $f(x)=\exp(x)$, hence, by defining property of $\exp$, $f(x)=\exp(x),\frac{\exp(x+y)}{\exp(y)}=\exp(x),\exp(x+y)=\exp(x)\exp(y)$.

It easily follows now that $\log(xy)=\log x+\log y$ for $x,y>0$. Now we can prove many properties of exponentiation which we are familiar with. Let me show this for $b^{x+y}=b^xb^y$ and $b^{xy}=(b^x)^y$.

$$b^{x+y}=\exp((x+y)\log b)=\exp(x\log b+y\log b)=\exp(x\log b)\exp(y\log b)=b^xb^y$$

$$b^{xy}=\exp(xy\log b)=\exp(y\log(\exp(x\log b)))=(\exp(x\log b))^y=(b^x)^y$$

Answer 2

From $\exp'(x)=\exp(x)$ we can find its Taylor series. We have $\exp^{(n)}(0)=1$ which implies that

$$\exp(x)=1+x+\tfrac1{2!}x^2+\tfrac1{3!}x^3+...$$

From this we find that $$\exp(x+y)=\exp(x)\cdot\exp(y)$$

Using these two properties we can show that for example $$2^2=\exp(2\ln 2)=\exp(\ln 2+\ln 2)=\exp(\ln 2)\cdot\exp(\ln 2)=2\cdot 2$$

For roots there's also an equivalence, consider $$2^{1/2}=\exp(\tfrac12\ln 2)=\exp(\ln 2-\tfrac12\ln 2)=\exp(\ln 2)\exp(-\tfrac12\ln 2)=\frac{2}{2^{1/2}}$$ $$2^{1/2}\cdot2^{1/2}=2$$

So $2^{1/2}$ is the number satisfying $$\prod_{n=1}^2 2^{1/2}=2$$

Answer 3

I will use a slightly different approach than Wojowu, who uses uniqueness in an essential way.

Let $g_y(x) = f(-x)f(x + y)$. Then $g'(x) = -f'(-x)f(x+y) + f(-x)f'(x+y) = 0$. Thus $g_y(x)$ is constant, with $g_y(x) = g_y(0) = f(0)f(y) = f(y)$. In particular, letting $y = 0$, we find $1 = f(0) = g_0(x) = f(-x)f(x)$. This shows that $f(x+y) = f(x)f(-x)f(x+y) = f(x)g_y(x)=f(x)f(y)$. This is the fundamental property of $f$.

The formula $f(x)f(-x) = f(0) = 1$ shows, in particular, that $f(x) \ne 0$. Since $f$ is continuous and $f(0) = 1$, the function must always be positive, by the intermediate value theorem. Since $f'(x) = f(x) > 0$, the function $f$ is strictly increasing from $\mathbb{R}$ to $(0,+\infty)$. Define $e = f(1) > f(0) = 1$.

The formula $f(nx) = f(x)^n$ (in which integral powers have their usual meaning) can be proved by induction for $n \geq 1$ using the fundamental formula, and extended to negative values of $n$ using $f(nx)f(-nx) = 1$. The fact that $f(n) = e^n$ shows $f(x)$ is unbounded below and above within $(0,+\infty)$, hence by the intermediate value theorem its range is all of $(0,+\infty)$. We now write $\exp$ and $\log$ for the function $f$ and its inverse.

Define $b^x$ to be $\exp(x \log b)$. When $x$ is an integer $n$, this has its usual meaning since $\exp(n \log b) = [\exp(\log b)]^n = b^n$. We have $\left(b^{1/q}\right)^q = \left[\exp\left(\frac{1}{q}\log b\right)\right]^q = \exp\left(q \cdot \frac{1}{q} \log b\right) = \exp(\log b) = b$. This proves that $b^{1/q} = \sqrt[q]{b}$. Now for integral $p$, we have $b^{p/q} = \exp\left(\frac{p}{q}\log b\right) = \left[\exp\left(\frac{1}{q}\log b\right)\right]^p = \left(b^{1/q}\right)^p = \left(\sqrt[q]{b}\right)^p$. Note also that $b^x$ is an increasing or decreasing bijection from $\mathbb{R}$ to $(0,+\infty)$, according to the sign of $\log b$, i.e., to whether $b > 1$ or $b< 1$. (The fact that the function $b^x$ has the expected values for rational $x$ and is monotonic characterizes it completely.)

We have $b^{x+y} = \exp[(x+y)\log b] = \exp(x \log b)\exp(y \log b) = b^x b^y$, from the fundamental formula.

We also have $(b^x)^y = \exp(y \log b^x) = \exp(y \log[\exp(x \log b)]) = \exp(yx \log b) = b^{xy}.$