About Higher Powers

This post isn’t about God; I write at length about God elsewhere.

We have explored expressions involving $x, x^2, x^3$ , and even binomials such as $(1+x)^n$ . Much of our exploration arises from interpreting powers as repeated multiplication in the following sense: given any positive number $a$ and positive integers $x, y$ ,

$a^{x+y} = a^x a^y, \quad (a^x)^y = a^{xy}.$

It turns out that we are allowed to regard $x, y$ as real numbers.

Theorem 1. Given any positive number $a > 0$ , for real numbers $x, y$ ,

$a^{x+y} = a^x a^y, \quad a^x > 0.$

We call $a^x$ the base– $a$ exponential of $x$ .

Proof. The complete construction of the exponential takes place here, using the tools in undergraduate real analysis.

Theorem 2 (Laws of Exponents). Using the exponential in Theorem 1, we have the following laws of exponents:

$\displaystyle a^0 = 1, \quad a^{1/n} = \sqrt[n]{a}, \quad a^{-x} = \frac 1{a^x}, \quad a^{x-y} = \frac{ a^x }{ a^y }.$

Here, $n$ refers to a positive integer, and $x,y$ refers to real numbers.

Proof. Using Theorem 1,

$a^0 = a^{0+0} = a^0 \cdot a^0 = (a^0)^2 \quad \Rightarrow \quad (a^0)^2 - a^0 = 0.$

Factorising $a^0 > 0$ ,

$a^0 (a^0 - 1) = (a^0)^2 - a^0 = 0 \quad \Rightarrow \quad a^0 - 1 = 0 \quad \Rightarrow \quad \boxed{ a^0 = 1 }.$

Similarly, by applying Theorem 1 a total of $n$ times,

$(a^{1/n})^n = \underbrace{a^{1/n} \cdot a^{1/n} \cdot \cdots \cdot a^{1/n}}_n = a^{n \cdot (1/n)} = a\quad \Rightarrow \quad \boxed{ a^{1/n} = \sqrt[n]{a} }.$

Using Theorem 1 and the first result,

$\displaystyle a^x \cdot a^{-x} = a^{x+(-x)} = a^0 = 1.$

Finally, using Theorem 1 and the previous result,

$\displaystyle a^{x-y} = a^{x + (-y)} = a^x \cdot a^{-y} = a^x \cdot \frac 1{a^y} = \frac{ a^x }{ a^y } \quad \Rightarrow \quad \boxed{ a^{x-y} = \frac{ a^x }{ a^y } }.$

Just like how we devoted nontrivial time and energy to solve equations that look like $x^2 = 2$ (it is $x = \pm \sqrt 2$ ), we would also like to solve equations such as $2^x = 3$ . We observe that, rather straightforwardly, if $k$ is an integer, then

$2^x = 2^k \quad \Rightarrow \quad x = k.$

However, how do we solve equations such as $2^x = 3$ ? The idea is to use better and better decimal approximations of $x$ , and we give the “perfect” answer—the base- $2$ logarithm of $3$ . For a proper, technical construction, see the same post pertaining exponential functions.

Definition 1. Given real constants $a, b$ with $a > 0$ , the base- $a$ logarithm of $b$ is the unique number $x$ such that

$a^x = b.$

In this case, we denote $x := \log_a(b)$ .

Theorem 3 (Inverse Property). Given real constants $a, b$ with $a > 0$ ,

$a^{\log_a(b)} = b,\quad \log_a(a^b) = b.$

Proof. Define $x := \log_a(b)$ . By Definition 1,

$a^{\log_a(b)} = a^x = b.$

Similarly, define $y := \log_a(a^b)$ . By Definition 1,

$a^y = a^b \quad \Rightarrow \quad \log_a(a^b) = y = b.$

Thanks to the basic property described in Theorem 1, the laws of exponents, and the inverse property, we recover all of the laws of logarithms. But to do that, we need our special friend $e \approx 2.718$ .

Theorem 4. Given positive constants $a, x, y$ , we have the following laws of logarithms:

$\begin{aligned} \log_a(xy) &= \log_a(x) + \log_a(y),\\ \log_a\left( \frac xy \right) &= \log_a(x) - \log_a(y). \end{aligned}$

Furthermore $\log_a(1) = 0,\log_a(a) = 1$ , and for any $r > 0$ ,.

$\log_a(x^r) = r \log_a(x).$

Proof Sketch. We prove the first property, prove a special case for the last property, and relegate the others as an exercise in definition chasing. Using Theorem 1 and the inverse property,

$a^{\log_a(x) + \log_a(y)} = a^{\log_a(x)} \cdot a^{\log_a(y)} = xy.$

By Definition 1, $\log_a(xy) = \log_a(x) + \log_a(y)$ . For the last property, we prove the special case that $r$ is a positive integer:

$\begin{aligned} \log_a(x^r) &= \log_a(\underbrace{ x \cdot x \cdot \cdots \cdot x }_r )\\ &= \underbrace{ \log_a(x) + \log_a(x) + \cdots + \log_a(x) }_r = r \log_a(x), \end{aligned}$

The general case uses our technical definition of the exponential as outlined in the real-analysis posts.

Corollary 1. Given real constants $a, x, y$ and $a > 0$ ,

$(a^x)^y = a^{xy}.$

Proof. Using the last property in the laws of logarithms,

$\log_a((a^x)^y) = y \log_a(a^x) = y \cdot x \log_a(a) = y \cdot x \cdot 1 = xy.$

By the inverse property,

$(a^x)^y = a^{\log_a((a^x)^y)} = a^{xy},$

as required.

Example 1. By considering the real number $r := {\sqrt 2}^{\sqrt 2}$ , demonstrate the existence of irrational numbers $a,b$ such that $a^b$ is rational. You may assume without proof that $\sqrt 2$ is irrational.

Solution. There are two cases to consider: either $r$ is rational or not. If $r$ is rational, then setting $a = b = \sqrt 2$ yields

$a^b = {\sqrt 2}^{\sqrt 2} = r,$

which is rational. If $r$ is irrational, then setting $a = r = {\sqrt 2}^{\sqrt 2}$ and $b = \sqrt 2$ yields, by Corollary 1,

$a^b = r^{\sqrt 2} = {\left( {\sqrt 2}^{\sqrt 2} \right)}^{\sqrt 2} = {\sqrt 2}^{\sqrt 2 \cdot \sqrt 2} = {\sqrt 2}^2 = 2,$

which is rational.

No matter which scenario is ultimately true, we can always arrive at the conclusion that there exist irrational numbers $a, b$ such that $a^b$ is rational.

Theorem 5 (Change-of-Base). Given positive numbers $a, b, c$ ,

$\displaystyle \log_a(b) = \frac{\log_c(b)}{\log_c(a)}.$

In particular, denoting $\lg \equiv \log_{10}$ and $\ln \equiv \log_e$ , where $e \approx 2.718$ denotes the exponential unit,

$\displaystyle \log_a(b) = \frac{\lg(b)}{\lg(a)} = \frac{\ln(b)}{\ln(a)}.$

Proof. Using the inverse property,

$\begin{aligned} \log_a(b) \cdot \log_c(a) &= \log_c(a^{\log_a(b)}) = \log_c(b) \quad \Rightarrow \quad \boxed{ \log_a(b) = \frac{\log_c(b)}{\log_c(a)} }. \end{aligned}$

The base- $10$ logarithm finds much use in numerically estimating the logarithms. For instance, the pH of a solution that measures the acidity of that solution (crucial in ecological applications) is defined by the number $-\lg([\mathrm H^+])$ , where $[\mathrm H^+]$ denotes the concentration of protons in that solution. The base- $e$ logarithm finds much use in theoretical analysis, since the rate of change of the function $\ln(x)$ is $1/x$ .

Exponentials and logarithms form the cornerstone of any discussion involving rates of change, i.e. calculus, and infect every serious field of study in STEM: population growth in biology, cooling laws in thermodynamics, logistic growth in machine learning, so on and so forth.

Perhaps these mathematical higher powers point to an ultimate higher power?

For now, let’s return to earth and return to practical mathematics by discussing measurements of various tangible objects.

—Joel Kindiak, 28 Oct 25, 1523H

KindiakMath

About Higher Powers

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About Higher Powers

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