Notice that if $q\in \Q$ and $a>0$ then $a^q = e^{\ln (a^q ) } = e^{q\ln a }$. This equation motivates the following definition.

Definition 9.4.1 For $a>0 $ and $x\in\R$, we define $a^x = e^{x\ln a}$. The function $f(x)=a^x $ is the exponential function with base $a$. Separately, we define $0^x =0 $ for $x>0 $. $\square$

Notice also that for $x\in \R$ and $a>0$, $\ln a^x = \ln (e^{x\ln a }) = x\ln a$. Hence, the power rule for the natural logarithm works even when the power is irrational.

We now show that the familiar rules for exponents are valid.

Theorem 9.4.2 For $x,y \in \R$ and $a,b>0$:

    a $a^{x+y} = a^x a^y$

    b $a^{x-y} = a^x/a^y$

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

    d $(ab)^x = a^x b^x$


(a) We compute: $a^{x+y} = e^{(x+y) \ln a } = e^{x\ln a + y\ln a} = e^{x\ln a } e^{y\ln a } = a^x a^y$.

The proof of (b) is similar and left as an exercise.

(c) We compute: $(a^x )^y = e^{y\ln(a^x)} = e^{yx\ln a} = a^{xy}$.

(d) We compute: $(ab)^x = e^{x\ln (ab) } = e^{x\ln a +x\ln b} = e^{x \ln a } e^{x\ln b} = a^x b^x$. $\qed$

Theorem 9.4.3 If $f(x) =a^x $ (with $a>0 $) then $f'(x) = a^x \ln a $.

Proof. $$f'(x) = {d\over dx}(e^{x\ln a}) = e^{x\ln a}\ln a =a^x \ln a.$$


Corollary 9.4.4 For $a>0 $ and $a\neq 1$, $\ds\int a^x\,dx = {a^x\over\ln a} + C$. $\qed$

We are now in a position to prove the general power rule.

Theorem 9.4.5 (Power Rule) If $f(x) = x^n $, $x>0$, and $n$ is any real number, then $f'(x) =nx^{n-1}$.

Proof. $\ds f'(x) ={d\over dx} x^n = {d\over dx} e^{n\ln x} = e^{n\ln x}{n\over x} = {nx^n \over x} = nx^{n-1}$. $\qed$

The restriction that $x>0 $ is necessary since we have not defined exponential expressions with negative bases and arbitrary real powers.

We now turn to logarithms base $a$. Note that if $a>0 $ and $a\neq 1 $ then $a^x \ln a \neq 0 $ for every $x$. Hence, the function $f(x) =a^x $ is injective.

Definition 9.4.6 If $a>0$ and $a\neq1$, the inverse of $a^x$ is called the logarithmic function base $a$. In symbols, we write this function as $\ds \log _a x$. $\square$

We exclude $a=1$ because $1^x = 1$ is not injective on any domain containing more than one point.

Remark If $a=10 $ we usually write $\log $ instead of $\log _{10}$, and of course $\log _e =\ln$. In more advanced texts, "$\log $'' refers to the natural logarithm.

Theorem 9.4.7 The following hold for $a, x, y >0 $, $a\neq 1 $, and $q\in \R$:

    a. $\ds\log _a (xy) =\log _a x +\log _a y $

    b. $\ds\log _a {x\over y} = \log _a x -\log _a y $

    c. $\ds\log _a x^q = q\log _a x .$

Proof. (a) Let $u= \log_a x$ and $v= \log_a y$. Then $a^u =x $ and $a^v = y$, and $\ds xy = a^u a^v = a^{u+v}$, so $\ds \log _a (xy) = u+v = \log_a x+\log_a y$.

The other parts are left as exercises. $\qed$

When computing decimal approximations to logs of arbitrary bases with a calculator or a computer algebra system the following result comes in handy.

Lemma 9.4.8 If $a, b>0 $, $a ,b \neq 1 $, and $x>0 $, < $\ds\log_a x = {\log_b x\over \log_b a}$.

Proof. Let $y=\log_a x $, so $a^y = x$. Then $\ds\log_b x = \log_b (a^y ) = y\log_b a =\log_a x \log _b a$. $\qed$

Typically this is useful when $b=e$ and $b=10$, since calculators can typically compute logarithms to those bases.

Theorem 9.4.9 $\ds{d\over dx} \log _a x = {1\over x\ln a}$.

Proof. By the preceding lemma, $f(x) =\ln x/\ln a$, and the derivative is then easy. $\qed$

Finally, we express $e^x$ as a limit. When $x=1$ we get a limit expression for $e$ which is sometimes taken as the definition of $e$.

Theorem 9.4.10 If $x\geq 0$, $\ds e^x = \lim_{n\to\infty} \left( 1+{x\over n}\right)^n$.

Proof. If $x=0$ both expressions are 1.

If $x>0$ we begin by rewriting the right side as we have before: $$\left(1+{x\over n}\right)^n=\left(e^{\ln (1+x/n)}\right)^n= e^{n\ln (1+x/n)}.$$ Now because $e^x$ is continuous, $$\lim_{n\to\infty}e^{n\ln (1+x/n)}=e^{\lim_{n\to\infty}n\ln (1+x/n)}.$$ So really we need to compute $\ds\lim_{n\to\infty}n\ln (1+x/n)$, for which we use L'Hôpitals rule: $$\lim_{n\to\infty}n\ln (1+x/n)=\lim_{n\to\infty}{\ln (1+x/n)\over 1/n}= \lim_{n\to\infty}{{1\over 1+x/n}{-x\over n^2} \over -1/n^2}= \lim_{n\to\infty}{1\over 1+x/n}x = x.$$ $\qed$

This same simple fact, $\ds a=e^{\ln a}$, is useful in many similar situations.

Example 9.4.11 Let $f(x) =x^x$, $x>0$. Compute $f'(x)$ and $\ds\lim_{x\to0^+} f(x)$.

Start with $f(x)=x^x = e^{x\ln x}$. Then $$f'(x)=e^{x\ln x}\left(x{1\over x}+\ln x\right)=x^x(1+\ln x).$$ For the limit, we again notice that $$\lim_{x\to0^+} x^x = \lim_{x\to0^+}e^{x\ln x} =e^{\lim_{x\to0^+} x\ln x}.$$ Then we compute the limit by L'Hôpital's rule again: $$\lim_{x\to0^+}x\ln x = \lim_{x\to0^+}{\ln x\over 1/x}= \lim_{x\to0^+}{1/x\over -1/x^2}=\lim_{x\to0^+}(-x)=0.$$ Thus $\ds\lim_{x\to0^+} x^x=e^0=1$. $\square$

Example 9.4.12 Compute $\ds\int_{\pi/6}^{\pi/3} 2^{\cos x} \sin x\; dx$.

Let $u=\cos x$, so $du=-\sin x\;dx $. Changing the limits, when $x=\pi/6$, $u =\sqrt{3}/2$, and when $x=\pi/3$, $u=1/2$. Then $$\int_{\pi/6}^{\pi/3} 2^{\cos x} \sin x\;dx =-\int_{\sqrt{3}/2}^{1/2} 2^u du = \left.-{2^u\over\ln 2}\right|_{\sqrt{3}/2}^{1/2} ={-2^{1/2} +2^{\sqrt{3}/2}\over\ln 2}.$$ $\square$

Exercises 9.4

Ex 9.4.1 Prove part (b) of theorem 9.4.2.

Ex 9.4.2 Sketch the graph of $\ds y=a^x $ in the three cases $a>1$, $a=1$, and $0< a< 1 $. What happens to the graph as $a\to 0^+$? What happens to the graph as $a\to\infty $?

Ex 9.4.3 Sketch the graph of $\ds y=\log_a x$ in the two cases $a>1$ and $0< a< 1 $. What happens to the graph as $a\to 0^+$? What happens to the graph as $a\to\infty$? (Use the previous exercise together with exercise 22 in section 9.1.)

Ex 9.4.4 Prove parts (b) and (c) of theorem 9.4.7.

Ex 9.4.5 Sketch the graph of $\ds y=3^{6x-1} + 5 $.

Ex 9.4.6 Sketch the graph of $\ds y= -(1/2)^{-3x}$.

Ex 9.4.7 Sketch the graph of $\ds y=4\log_2 (12x + 6)-2$.

Ex 9.4.8 Compute the second derivative of $\ds f(x)=x^x$.

Ex 9.4.9 Compute $\ds f'(\pi/4)$ when $\ds f(x) = 5^{\sin 3x} + \log_7 x$.

Ex 9.4.10 Compute the derivative of $\ds f(x) = 3^x - 4x^2 + \sin(3x) - \pi^e$.

Ex 9.4.11 Compute $\ds \int_1^2 3^x - x^3\,dx$.

Ex 9.4.12 Compute $\int \sin (2^x ) 2^x dx $.

Ex 9.4.13 Find the area of the region given by $\ds \{(x, y) \mid 1 \leq x \leq 2 , 2^x \leq y \leq 3^x\}$.

Ex 9.4.14 Find the average of the function $\ds f(x)= 2x\cdot 5^{x^2}$ on the interval $[4,9]$.

Ex 9.4.15 Find the volume of the solid obtained by rotating the region $\ds \{(x,y) \mid 2 \leq x \leq 4, (\log _2 x)/x \leq y \leq 2^x \}$ about the line $y=-1 $.

Ex 9.4.16 Show that $\log_a x =-\log_{1/a} x $ for any $a>0, a\neq 1$. Interpret this result geometrically; that is, sketch the graph of $y=\log_a x$ and $y=\log_{1/a} x$ on the same diagram and point out how the graphs are related to each other.