How do I prove exponent rules for all real numbers?

Question

It is easy to prove exponent rules for only the positive integers. For example, a^m.a^n = (a.a.a.a...)m times . (a.a.a.a.a....)n times so we just add the exponents m and n to get a^m+n. But how do I apply the same logic on rational or irrational or negative exponents ? How do I prove that these laws of exponents work for all real numbers NOT only just the positive integers ? for example (a.b)^n = a^n.b^n or any other law of exponents.

Answer 1

In my earlier answer "Proof of derivative of xⁿ for real n without binomial expansion?" the exponent (index) $n$ is generalized from positive integers to zero to negative integers to rationals to reals—the last step being either less rigorous or more difficult. An alternative (noted in the comments on that question, and elsewhere) is to define $x^\alpha$ as $\exp(\alpha\ln x)$ from the get-go. This definition has the advantage that it works for all real $\alpha$, but the disadvantage that the get-go takes some work to get going: to avoid circular argument, we must first define $\ln$ and $\exp$ and prove some of their properties without using powers, and then use those results to prove whatever properties of powers we are interested in—including any properties that we previously regarded as definitions.

But the amount of work is not prohibitive for a StackExchange answer. So, encouraged by the apparent lack of a "readable, self-contained account of all the details required" (quoting Timothy Chow on the present page), I offer the following.

Powers with real exponents

Without any prior notion of powers, we can establish \begin{align} \tfrac{d}{dx}\big(\alpha f(x)\big) &= \alpha\tfrac{d}{dx}\big(f(x)\big)\!\!\! &&\text{for constant}~\alpha;\\[1ex] \tfrac{d}{dx}\Big(y\big(u(x)\big)\Big) & = \frac{dy}{du}~\!\frac{du}{dx} && \text{[chain rule]};\\[1ex] \tfrac{d}{dx} \int_a^x \!\!f(t)\,dt \, &= f(x) && [\text{first fundamental theorem}]. \end{align} And we can think of $dt$ (in the last equation) as an infinitesimal change in the abscissa, and $f(t)$ as the ordinate, and the integral as the summation of the rectangular areas $f(t)~\!dt$  over the indicated range of $t$, giving the total area.

Definition of  $\boldsymbol{\ln}$:  For present purposes, let us define $$\ln x \,:= \int_1^x \!\tfrac{\,1\,}{t}\,dt \,.$$ This definition doesn't assume that $\ln$ is any sort of logarithm (whatever that means). But it does imply (by putting ${x\!=\!1}$) that $$\ln(1) = 0 \,.$$ It also implies (by the first fundamental theorem) that $$\tfrac{d}{dx}\big(\!\ln x\big) = \tfrac{\,1\,}{x} \,.$$ So the $\ln$ function is a primitive (antiderivative) of the reciprocal function, with the initial value  ${\ln(1)\!=\!0}$.

Domain of  $\boldsymbol{\ln}$:  It is clear from the definition that  $\ln x$  exists for all  ${x\!>\!0}$  (i.e., ${x\in\mathbb{R}_*^+}$) and is an increasing function on that domain.

Product rule for $\boldsymbol{\ln}$:  By definition, \begin{align} \ln(ab)\, &= \int_1^{ab} \!\tfrac{\,1\,}{t}\,dt && \\ &= \int_1^a \!\tfrac{\,1\,}{t}\,dt \,+ \int_a^{ab} \!\tfrac{\,1\,}{t}\,dt && [\text{splitting the area}]\\ &= \,\ln a \,+ \int_a^{ab} \!\tfrac{\,1\,}{t}\,dt \,. && \end{align} In the remaining integral, if we put ${t\!=\!au}$, then ${dt\!=\!a~\!du}$ and, as $t$ ranges from $a$ to $ab$,  $u$ ranges from $1$ to $b$.  So the remaining integral becomes $$\int_1^{b} \!\tfrac{\,1\,}{au}\,a\,du \,= \int_1^{b} \!\tfrac{\,1\,}{u}\,du \,=:\, \ln b$$ and we have $$\boldsymbol{\ln(ab) = \ln a + \ln b \,.}$$ This rule can be extended to any number of factors: \begin{align} \ln(abc) &= \ln a + \ln(bc)\\ &= \ln a + \ln b + \ln c \,, \end{align} and so on.

Range of  $\boldsymbol{\ln}$:  It is clear from the definition that the range of the function  $\ln x$ is continuous and includes  $\ln 2$,  which is positive, and  $\ln\big(\frac{\,1\,}{2}\big)$,   which is negative. But how wide is the range? By the product rule just found, \begin{align} \ln(2\!\times\!2) &= \ln 2 + \ln 2 \,;\\ \ln(2\!\times\!2\!\times\!2) &= \ln 2 + \ln 2 + \ln 2 \,; \\ \ln(2\!\times\!2\!\times\!2\!\times\!2) &= \ln 2 + \ln 2 + \ln 2 + \ln 2 \,;\\ & \text{etc.} \end{align} Continuing in this manner, we can make the right-hand side arbitrarily large and positive by making the argument of  $\ln$ sufficiently large and positive. Similarly, \begin{align} \ln\big(\tfrac{\,1\,}{2}\!\times\!\tfrac{\,1\,}{2}\big) &= \ln\!\big(\!\tfrac{\,1\,}{2}\!\big) + \ln\!\big(\!\tfrac{\,1\,}{2}\!\big) \,;\\ \ln\big(\tfrac{\,1\,}{2}\!\times\!\tfrac{\,1\,}{2}\!\times\!\tfrac{\,1\,}{2}\big) &= \ln\!\big(\!\tfrac{\,1\,}{2}\!\big) + \ln\!\big(\!\tfrac{\,1\,}{2}\!\big) + \ln\!\big(\!\tfrac{\,1\,}{2}\!\big) \,; \\ \ln\big(\tfrac{\,1\,}{2}\!\times\!\tfrac{\,1\,}{2}\!\times\!\tfrac{\,1\,}{2}\!\times\!\tfrac{\,1\,}{2}\big) &= \ln\!\big(\!\tfrac{\,1\,}{2}\!\big) + \ln\!\big(\!\tfrac{\,1\,}{2}\!\big) + \ln\!\big(\!\tfrac{\,1\,}{2}\!\big) + \ln\!\big(\!\tfrac{\,1\,}{2}\!\big) \,;\\ & \text{etc.} \end{align} Continuing in this manner, we can make the right-hand side arbitrarily large and negative by making the argument of  $\ln$ sufficiently small and positive. In summary,  $\ln x$  has no upper bound as  ${x\!\to\!+\infty}$,  and no lower bound as  ${x\!\to\!0^+}$.  As its domain indeed extends from $0^+$ to $+\infty$  and its range is continuous, its range is the whole of  $\mathbb{R}$.

Definition of $\boldsymbol{\exp}$:  As the $\ln$ function is increasing with domain $\mathbb{R}_*^+$ and range $\mathbb{R}$, it has an inverse function which is increasing with domain $\mathbb{R}$ and range $\mathbb{R}_*^+$. Let us call this inverse "$\exp$". Then $$\boldsymbol{\exp\ln x = \ln\exp x = x \,.}$$ We do not (yet) claim that $\exp$ is any sort of power function. But we do already know that  ${\ln(1)\!=\!0}$, whence the inverse relation gives $$\exp(0) = 1 \,.$$

Derivative of $\boldsymbol{\exp}$:  Recall that ${\frac{d}{dx}\big(\!\ln x\big) = \frac{\,1\,}{x}}$.  If  ${y = \exp x}$,  then  ${\ln y = x}$.  Differentiating both sides w.r.t. $x$, using the chain rule on the left, gives  ${\frac{\,1\,}{y}\frac{dy}{dx} = 1}$,  whence  ${\frac{dy}{dx} = y}$;  that is, $$\boldsymbol{\tfrac{d}{dx}\big(\!\exp x\big) = \exp x \,.}$$ So the $\exp$ function is its own derivative and its own primitive, with the initial value  ${\exp(0)\!=\!1}$.

Sum rule for $\boldsymbol{\exp}$:  By the product rule for $\ln$, \begin{align} \ln\big(\!\exp\alpha\,\exp\beta\big) &= \ln\exp\alpha+\ln\exp\beta \\ &= \alpha+\beta \,. \end{align} Hence, by the inverse relation (or by taking $\exp$ of both sides), $$\boldsymbol{\exp(\alpha\!+\!\beta) = \exp\alpha\,\exp\beta \,.}$$ This rule can be extended to any number of terms: \begin{align} \exp\big(\alpha\!+\!\beta\!+\!\gamma\big) &= \exp\alpha\,\exp\big(\beta\!+\!\gamma\big) \\ &= \exp\alpha\,\exp\beta\,\exp\gamma \,, \end{align} and so on.

Now we're ready for the main item of business:  Define the function $$P_{\alpha}(x) := \exp(\alpha\ln x) \qquad [\text{for}~{x\!>\!0}].$$ By that definition we have \begin{align} P_{\alpha+\beta}(x) &= \exp\big((\alpha\!+\!\beta)\ln x\big) \\ &= \exp\big(\alpha\ln x + \beta\ln x\big) \\ &= \exp(\alpha\ln x)~\!\exp(\beta\ln x) \,, \end{align} i.e. $$P_{\alpha+\beta}(x) = P_{\alpha}(x) ~\!P_{\beta}(x) \,;$$ this is a sum rule for $\boldsymbol{P}$, which again can be extended to any number of terms. Putting ${\alpha\!=\!1}$ in the same definition gives ${P_1(x) = \exp\ln x}$, i.e. $${P_1(x) = x \,.}$$ Knowing $P_1(x)$ and repeatedly applying the sum rule for $P$, we get \begin{equation}\begin{split} P_2(x) &= P_{1+1}(x)=P_1(x)\!\cdot\!P_1(x)=x\cdot x ~; \\ P_3(x) &= P_{2+1}(x)=P_2(x)\!\cdot\!P_1(x)=x\cdot x\cdot x ~; \\ P_4(x) &= P_{3+1}(x)=P_3(x)\!\cdot\!P_1(x)=x\cdot x\cdot x\cdot x ~;\\[1ex] & \text{etc.} \end{split}\end{equation}

Definition of $\boldsymbol{x^{\alpha}}$:  At this point it is safe to rename $\boldsymbol{P_{\alpha}(x)}$ as $\boldsymbol{x^{\alpha}}$,  so that the last definition and the subsequent results can be rewritten as \begin{equation}\label{grpe1}\tag{1} x^{\alpha} := \exp(\alpha\ln x) \quad [\text{for}~{x\!>\!0}] \,, \end{equation} \begin{equation}\label{grpe2}\tag{2} x^{\alpha+\beta} = x^{\alpha}~\!x^{\beta} \end{equation} (extendable to any number of terms), \begin{equation}\label{grpe3}\tag{3} x^1 = x \,, \end{equation} and \begin{equation} \begin{split} x^2 &= x\cdot x ~, \\ x^3 &= x\cdot x\cdot x ~, \\ x^4 &= x\cdot x\cdot x\cdot x ~, \\[1ex] & \text{etc.,} \end{split}\label{grpe4}\tag{4} \end{equation} where (\ref{grpe1}) and (\ref{grpe2}) are valid for real $\alpha$ & $\beta$, and (\ref{grpe4}) depends solely on (\ref{grpe2}) and (\ref{grpe3}).

I say "it is safe" because if the reader is carrying any other competing notions of "powers", these too will be obtainable directly or indirectly from (\ref{grpe2}) and/or (\ref{grpe3}).  E.g., \begin{align} x^{\alpha} &= x^{\alpha+0} && \\ &= x^{\alpha}\;\!x^0 && [\text{by}~(\ref{grpe2})] \end{align} and this is an identity, so we must have \begin{equation}\label{grpe5}\tag{5} x^0 = 1 \end{equation} [which also follows directly from definition (\ref{grpe1}) for ${x\!>\!0}$].  And \begin{align} x^{\alpha}\;\!x^{-\alpha} &= x^{\alpha-\alpha} && [\text{by}~(\ref{grpe2})]\\ &= x^0 && \\ &= 1 && [\text{by}~(\ref{grpe5})], \end{align} so that \begin{equation}\label{grpe6}\tag{6} x^{-\alpha} = \frac{1}{\,x^{\alpha}} \,. \end{equation} If $m$ and $n$ are both positive integers, it is clear from (\ref{grpe3}) and (\ref{grpe4}) that \begin{equation}\label{grpe7}\tag{7} x^{mn} = \big(x^m\big)^n , \end{equation} because this is merely grouping the factors. And if $m$ is real while $n$ is an integer greater than $1$, then, by the extended version of (\ref{grpe2}), $$x^{m/n}\cdot x^{m/n}\cdots~[\text{to}~n~\text{factors}] = x^m ,$$ so that \begin{equation}\label{grpe8}\tag{8} x^{m/n} = \sqrt[n]{x^{m\mathstrut}} \,. \end{equation}

One advantage of working from definition (\ref{grpe1}), of course, is that it circumvents the integer restrictions in (\ref{grpe7}) and (\ref{grpe8}). For (\ref{grpe7}), if $m$ and $n$ are real, \begin{align} \big(x^m\big)^n &= \exp\Big(n\ln\big(x^m\big)\Big) && [\text{by}~(\ref{grpe1})] \\ &= \exp\Big(n\ln\big(\!\exp(m\ln x)\big)\Big) && [\text{by}~(\ref{grpe1})~\text{again}] \\ &= \exp\Big(n(m\ln x)\Big) && [\text{since}~\ln~\text{inverts}~\exp] \\ &= \exp\big(mn\ln x\big) && \\[1ex] &= x^{mn} , && \end{align} as expected. This in turn implies that $x^{m/n}$ is the $n$th root of $x^m$, in the sense that $$\big(x^{m/n}\big)^n =~\! x^{(m/n)n} = x^m .$$ Another advantage is that because (\ref{grpe1}) implies (\ref{grpe2}) for all real exponents, it ensures that (\ref{grpe2}) remains correct for all subsequent interpretations of those exponents. If, instead, we start with (\ref{grpe2}) for positive integer exponents, and then apply it to exponents $\alpha$ and $-\alpha$ to derive (\ref{grpe6}), we have only found a necessary condition for negative indices to satisfy (\ref{grpe2}); we have not yet shown that this condition is sufficient for all possible combinations of positive and negative integer indices. Then a similar issue arises when we extend (\ref{grpe2}) to rational indices in deriving (\ref{grpe8}). But we have avoided these pitfalls by starting from (\ref{grpe1}).

The disadvantage of (\ref{grpe1}) is that it requires $x$ to be positive in order to keep everything within the real numbers—whereas, if all the indices are integers, equations (\ref{grpe2}) to (\ref{grpe7}) make sense if $x$ is negative; and if the indices are non-negative integers, equations (\ref{grpe2}) to (\ref{grpe5}) and (\ref{grpe7}) make sense even if ${x\!=\!0}$. But, for negative $x$, equation (\ref{grpe8}) causes difficulties: it works for even $m$, but gives no real result for odd $m$ and even $n$, but works again if we then double both $m$ and $n$, and it gives a negative result if $m$ and $n$ are both odd, but a positive result if we then double both $m$ and $n$ (unless we admit two roots). To avoid such paradoxes, we conventionally require the base to be positive if the index is a non-integer.

Reciprocal rule for $\boldsymbol{\ln}$:  By the product rule for $\ln$, $$\ln\tfrac{\,1\,}{x} + \ln x = \ln\big(\!\tfrac{\,1\,}{x}x\big) = \ln(1) = 0 \,,$$ whence $$\boldsymbol{\ln\tfrac{\,1\,}{x} = -\ln x \,.}$$

Definition of $\boldsymbol{x^{\alpha}}$:  Now we come to the question that brought me here:  What is ${\frac{d}{dx}\big(x^{\alpha}\big)}$ for real $\alpha\;\!$?  By definition (\ref{grpe1}), \begin{align} \tfrac{d}{dx}\big(x^{\alpha}\big) &= \tfrac{d}{dx}\big(\!\exp(\alpha\ln x)\big) && \\ &= \exp(\alpha\ln x)\tfrac{d}{dx}\big(\alpha\ln x \big) && [\text{chain rule}]\\ &= \exp(\alpha\ln x)\,\alpha~\!\tfrac{d}{dx}\big(\!\ln x \big) && \\ &= \exp(\alpha\ln x)\,\alpha~\!\tfrac{\,1\,}{x} && \\ &= \alpha\,\exp(\alpha\ln x)~\!\tfrac{\,1\,}{x} && \\ &= \alpha\,\exp(\alpha\ln x)~\!\exp\big(\!\ln\tfrac{\,1\,}{x}\big) && [\text{inverse functions}]\\ &= \alpha\,\exp\big(\alpha\ln x + \ln\tfrac{\,1\,}{x}\big) && [\text{sum rule for}~\exp]\\ &= \alpha\,\exp\big(\alpha\ln x - \ln x\big) && [\text{reciprocal rule}]\\ &= \alpha\,\exp\big((\alpha\!-\!1)\ln x\big) \,; && \end{align} i.e., by the same definition, $$\boldsymbol{\tfrac{d}{dx}\big(x^{\alpha}\big) = \alpha x^{\alpha-1} .}$$ While we're at it, let's see what happens if $x$ is the exponent: \begin{align} \tfrac{d}{dx}\big(a^x\big) &= \tfrac{d}{dx}\big(\!\exp(x\ln a)\big) && \\ &= \exp(x\ln a) \ln a && [\text{chain rule}];\\ \end{align} i.e., $$\tfrac{d}{dx}\big(a^x\big) = a^x \ln a \,.$$ So, for the function $a^x$,  differentiation w.r.t. $x$ reduces to multiplication by a scale factor, namely $\ln a$;  we say that $a^x$ is an eigenfunction of the $\frac{d}{dx}$ operator, and that  $\ln a$  is the corresponding eigenvalue of the operator.

Euler's number / Napier's constant:  Finally, as if it were an afterthought, let us define $$e := \exp(1)$$ or, equivalently, $$\ln e = 1 \,.$$ Then, by definition (\ref{grpe1}), $$e^x = \exp(x\ln e) = \exp(x\;\!1) \,;$$ i.e., $$\exp x = e^x .$$ Hence, by the usual definition of logarithms, the inverse of  $\exp$  is the logarithm to the base $e$; that is, $$\ln x = \log_e x \,.$$ Moreover, by definition (\ref{grpe1}) and the definition of  $\ln$, $$\big(1\!+\!\tfrac{\,x\,}{N}\big)^N =~\! \exp\Big(N\ln\big(1\!+\!\tfrac{\,x\,}{N}\big)\Big) = \exp\Big(N\!\int_1^{1+x/N} \!\tfrac{\,1\,}{t}~\!dt\Big) \,.$$ In the integrand,  $t$ is bounded by $1$ and ${1\!+\!x/N}$. Hence, as ${N\!\to\!\infty}$,  we have  ${t\!\to\!1}$,  so that $$\lim_{N\to\infty}\!\Big\{\big(1\!+\!\tfrac{\,x\,}{N}\big)^N\Big\} = \exp\!\Big(\!N\!\int_1^{1+x/N} \!1~\!dt\Big) = \exp\!\big(N~\!\tfrac{\,x\,}{N}\big) = \exp x \,;$$ i.e., $$e^x = \lim_{N\to\infty}\Big\{\big(1\!+\!\tfrac{\,x\,}{N}\big)^N\Big\} \,.$$ In particular (for ${x\!=\!1}$), $$e~\!= \lim_{N\to\infty}\Big\{\big(1\!+\!\tfrac{\,1\,}{N}\big)^N\Big\} \,.$$

 

If I've forgotten anything, take me to task in the comments!

Answer 2

The short answer, as already mentioned in a comment, is to begin by defining the exponential function by $$\exp(x) = 1 + x + {x^2\over 2!} + {x^3\over 3!} + {x^4 \over 4!} + \cdots$$ for any real (or even complex) $x$. Then $\log(x)$ is defined to be the inverse of $\exp(x)$, and exponentiation is defined by $$ a^b := \exp(b \log(a)).$$ The rules of exponents can then be proved by reasoning with the power series of the exponential function.

Contrary to what some commenters have suggested, I do not think that it is easy to find a readable, self-contained account of all the details required to flesh out the above sketch. I would suggest starting with the Prologue to Walter Rudin's book Real and Complex Analysis. This is rather terse but it does at least sketch the main steps.

Another approach, which avoids the exponential function, is sketched in Exercise 6 of Chapter 1 of Rudin's book Principles of Mathematical Analysis. This approach first carefully defines what $a^n$ (and therefore $a^{1/n}$) means for any real $a$ and any positive integer $n$. Then it defines $a^b$ to be the least upper bound of all numbers $a^t$ where $t$ is a rational number less than $b$. The advantage of this approach is that you can see more explicitly how to build up everything step by step from the axioms for the real numbers. But it's the definition in terms of the exponential function that is ultimately more useful in practice.

Answer 3

Claim: Given $n\in N, x,c\in R, c>0, x^n=c$ has at least 1 solution over the reals.

Let $A=\{x\in R| x^n\le c\}$

$\forall n, 0^n=0<c \implies 0\in A$.

If $c=1$, $x=1$ and we have our solution.

Let $x,y \in R, 0<x<y. $

$x^2<xy.$

$xy<y^2$

So $x^2<y^2$.

More generally $x<y\implies x^n<y^n$.

If $c<1$ and $x<c$, $x^n<x<c<1$ so $A$ is bounded above by 1.

If $1<c, c<x, c<c^n<x^n, so \ x\notin A.$ So $x\le c$. So $A$ is bounded above by $c$.

So either $c=1$ or A is bounded above by 1 or $c$. By the Least Upper Bound property of the reals a non-empty subset that is bounded above has a least upper bound. The least upper bound of A is the root we seek. And it thus makes sense to let $x=c^{1/n}$. From there we have $m\in N, x^m=c^{m/n}$ by the usual exponent rules. We now have a definition for raising to a rational power.

By the Archimidean Property $\forall r\in R, n \in N, \exists m/n \in Q| m/n<r<(m+1)/n$.

$n$ can be made arbitrarily small and whatever value we assign to $a^r$ it's difference from $a^{m/n}$ can be made arbitrarily small. This allows assigning it a value of the limit of a base raised to a sequence of rational numbers.