2.3 The Exponential and Natural Logarithm Functions

Functions of the form $a^x$, where the exponent $x$ varies, are called exponential functions. Unless otherwise noted, assume that $a > 0$ ($0^x$ is just 0, and $(-1)^{1/2}$ is not a real number). You already know how $a^x$ is defined when the exponent $x$ is a rational number (i.e. $x = m/n$ where $m$ and $n$ are integers, $n \ne 0$). But what if $x$ were irrational, such as $\sqrt{2}$? What would $3^{\sqrt{2}}$ mean?

The idea is that $\sqrt{2} = 1.414213562\ldots$ can be approximated by rational numbers $14/10 = 1.4$, $141/100 = 1.41$, $1414/1000 = 1.414$, $14142/10000 = 1.4142$, and so on, taking larger and larger numerators and denominators in the rational approximations to get more and more decimal places of $\sqrt{2}$. Then $3^{\sqrt{2}}$ would be the number that 3 raised to those rational approximations approaches:

\[ 3^{\sqrt{2}} = \lim_{\genfrac{}{}{0pt}{}{m/n \to \sqrt{2}}{m/n~\text{rational}}}\;3^{m/n} \]

Each quantity $3^{m/n}$ is defined inside the above limit, and as the rational numbers $m/n$ get closer to the value of $\sqrt{2}$ it can be shown the limit of the values of $3^{m/n}$ will exist:^[1]

\begin{alignat*}{3} 3^{1.4} ~&=~ 3^{\frac{14}{10}} ~&=~ 4.65553672174608\\ 3^{1.41} ~&=~ 3^{\frac{141}{100}} ~&=~ 4.70696500171657\\ 3^{1.414} ~&=~ 3^{\frac{1414}{1000}} ~&=~ 4.72769503526854\\ 3^{1.4142} ~&=~ 3^{\frac{14142}{10000}} ~&=~ 4.72873393017119\\ 3^{1.41421} ~&=~ 3^{\frac{141421}{100000}} ~&=~ 4.72878588090861\\ 3^{1.414213} ~&=~ 3^{\frac{1414213}{1000000}} ~&=~ 4.72880146624114\\ {} &\vdots \quad ~&\vdots \\ 3^{1.414213562\ldots} ~&=~ 3^{\sqrt{2}} ~&=~ 4.72880438783742 \end{alignat*}

Of course you would never do all this by hand—you would simply use a computer or calculator, which use much more efficient algorithms for calculating powers in general.^[2]

All the usual rules of exponents that you learned in algebra apply to $a^x$ when defined in the manner described above, with $a > 0$ and $x$ varying over all real numbers. Of all the possible values for the base $a$, the one that appears the most in mathematics, the sciences and engineering is the base $e$, defined as:

Definition 2.1

\begin{equation} e ~=~ \lim_{x \to \infty} ~\left( 1 ~+~ \frac{1}{x} \right)^x \tag{2.1}\end{equation}

The approximate value of is $e ~=~ 2.71828182845905\ldots$ (often called the Euler number).

The limit in this definition means that as $x$ becomes larger—approaching infinity ($\infty$)—the values of $\left( 1 ~+~ \frac{1}{x} \right)^x$ approach a number, denoted by $e$. More decimal places for $e$ can be obtained by making $x$ sufficiently large.^[3] For example, when $x=5 \times 10^6$ the value is 2.718281555200129. For extremely large values of $x$, that is, when $x \gg 1$ (the symbol $\gg$ means “much larger than”),

\[e ~\approx~ \left( 1 ~+~ \frac{1}{x} \right)^x \quad\Rightarrow\quad e^{1/x} ~\approx~ \left(\left( 1 ~+~ \frac{1}{x} \right)^x\right)^{1/x} ~=~ 1 ~+~ \frac{1}{x} \quad\Rightarrow\quad \left(e^{1/x} ~-~ 1\right)\,x ~\approx~ \left(\frac{1}{x}\right)\,x ~=~ 1 ~,\]

so letting $h = 1/x$, and noting that $h = 1/x \to 0$ if and only if $x \to \infty$, yields the useful limit:^[4]

Theorem 2.5

\begin{equation} \lim_{h \to 0}~\frac{e^h ~-~ 1}{h} ~=~ 1 \tag{2.2}\end{equation}

Using the above limit, the derivative of $y = e^x$ can be found:

Theorem 2.6

\[ \ddx\,\left(e^x\right) ~=~ e^x \]

proof[Proof:] Using the limit definition of the derivative for $f(x) = e^x$,

\begin{align*} \ddx\,\left(e^x\right) ~&=~ \lim_{h \to 0} ~\frac{f(x+h) ~-~ f(x)}{h} ~=~ \lim_{h \to 0} ~\frac{e^{x+h} ~-~ e^x}{h}\\[6pt] &=~ \lim_{h \to 0} ~\frac{e^x \left(e^h ~-~ 1\right)}{h} ~=~ e^x \;\cdot\; \lim_{h \to 0}~\frac{e^h ~-~ 1}{h} \quad\text{(since $e^x$ does not depend on $h$)}\\[6pt] &=~ e^x \;\cdot\; 1 ~=~ e^x \end{align*}

In general, for a differentiable function $u = u(x)$ as the exponent the Chain Rule yields:

Theorem 2.7

\[ \ddx\,\left(e^u\right) ~=~ e^u \;\cdot\; \dudx \]

Example 2.5

Find the derivative of $y = 4e^{-x^2}$.

Solution: $\dydx ~=~ 4 e^{-x^2} \;\cdot\; \ddx\,(-x^2) ~=~ 4 e^{-x^2} \;\cdot\; (-2x) ~=~ -8x e^{-x^2}$

The function $e^x$ is often referred to simply as the exponential function, even though there are obviously many exponential functions. What makes the base $e$ so special? Take $y = Ae^{kt}$ to represent the amount of some physical quantity at time $t$, for some constants $A$ and $k$. Then

\[\dydt ~=~ \ddt\,\left(Ae^{kt}\right) ~=~ k \cdot Ae^{kt} = ky ~,\]

which says that the instantaneous rate of change of the quantity is directly proportional to the amount present at that instant. It turns out that many physical quantities exhibit that behavior, some of which will be discussed shortly. Conversely, in Chapter 5 it will be shown that any solution to the differential equation $\dydt = ky$ must be of the form $y = Ae^{kt}$ for some constant $A$. This is what gives the exponential function its special significance.

Let $f(x) = e^x$ be the exponential function. Then $f(x) > 0$ for all $x$ and $f'(x) = f(x) = e^x > 0$ for all $x$, and so $f(x)$ is strictly increasing. The graph is shown in Figure 2.3.1.

tikzpicture[>=latex,every node/.style=font=,domain=-1.9:3.9] [->,black!60,line width=1pt] (-2,0) -- (4,0) node[pos=1.0,right] x; [->,black!60,line width=1pt] (0,0) -- (0,4) node[pos=1.0,above] y; [below] at (0,0) 0; [above left] at (0,1) 1; [linecolor,line width=1.5pt] plot ( ,exp(0.346* )); [below right] at (2,2) y = e^x; (0,1) circle (2.5pt); tikzpicture — Figure 2.3.1The exponential function

tikzpicture[>=latex,every node/.style=font=,domain=0.585:4.9] [->,black!60,line width=1pt] (-1,0) -- (5,0) node[pos=1.0,right] x; [->,black!60,line width=1pt] (0,-1) -- (0,3.45) node[pos=1.0,above] y; [below left] at (0,0) 0; [below right] at (1,0) 1; [linecolor,line width=1.5pt] plot ( ,1.864*ln( )); [above left] at (2.5,2) y = \,x; (1,0) circle (2.5pt); tikzpicture — Figure 2.3.2Natural logarithm function

Thus, the exponential function is one-to-one over the set of all real numbers and hence has an inverse function, called the natural logarithm function, denoted (as a function of $x$) as $f^{-1}(x) = \ln\,x$. The graph is shown in Figure 2.3.2. Below is a summary of the relationship between $e^x$ and $\ln\,x$:

\begin{alignat*}{3} \text{domain of $\ln\,x$} \quad&=\quad \text{all $x > 0$} \quad&&=\quad \text{range of $e^x$}\\ \text{range of $\ln\,x$} \quad&=\quad \text{all $x$} \quad&&=\quad \text{domain of $e^x$} \end{alignat*}

\begin{gather} y ~=~ e^x \quad\text{if and only if}\quad x ~=~ \ln\,y\tag{2.3}\\ e^{\ln\,x} ~=~ x \quad\text{for all $x > 0$}\tag{2.4}\\ \ln\,\left(e^x\right) ~=~ x \quad\text{for all $x$} \tag{2.5}\end{gather}

The reader should be aware that many—if not most—fields outside of mathematics use the notation $\log\,x$ instead of $\ln\,x$ for the natural logarithm function.^[5] From algebra you should be familiar with the following properties of the natural logarithm, along with their equivalent properties in terms of the exponential function:^[6]

\begin{align*} \ln\,(a b) ~&=~ \ln\,a ~+~ \ln\,b & e^{a} \cdot e^{b} ~&=~ e^{a + b}\\ \ln\,\left(\frac{a}{b}\right) ~&=~ \ln\,a ~-~ \ln\,b & \frac{e^a}{e^b} ~&=~ e^{a - b}\\ \ln\,a^b ~&=~ b\, \ln\,a & \left(e^a\right)^b ~&=~ e ^{ab}\\ \ln\,1 ~&= 0 & e^0 ~&=~ 1 \end{align*}

To find the derivative of $y = \ln\,x$, use $x = e^y$:

\[\dydx ~=~ \frac{1}{\dxdy} ~=~ \frac{1}{\ddy\,\left(e^y\right)} ~=~ \frac{1}{e^y} ~=~ \frac{1}{x}\]

Hence:

Theorem 2.8

\[ \ddx\,\left(\ln\,x\right) ~=~ \frac{1}{x} \]

In general, for a differentiable function $u = u(x)$, the Chain Rule yields:

Theorem 2.9

\[ \ddx\,\left(\ln\,u\right) ~=~ \frac{1}{u} \;\cdot\; \dudx ~=~ \frac{u'}{u} \]

Example 2.6

Find the derivative of $y = \ln\,\left(x^2 + 3x - 1\right)$.

Solution: $\Dydx ~=~ \dfrac{1}{x^2 + 3x - 1} \;\cdot\; \Ddx\,(x^2 + 3x - 1) ~=~ \dfrac{2x + 3}{x^2 + 3x - 1}$

Recall that $\abs{x} = -x$ for $x < 0$, in which case $\ln\,(-x)$ is defined and

\[ \ddx\,(\ln\,\abs{x}) ~=~ \ddx\,(\ln\,(-x)) ~=~ \frac{1}{-x} \;\cdot\; (-1) ~=~ \frac{1}{x} ~. \]

Combine that result with the derivative $\ddx\,(\ln\,x) = \frac{1}{x}$ for $x > 0$ to get:

Theorem 2.10

\[ \ddx\,(\ln\,\abs{x}) ~=~ \frac{1}{x} \]

Logarithmic Differentiation

For some functions it is easier to differentiate the natural logarithm of the function first and then solve for the derivative of the original function. This technique is called logarithmic differentiation, demonstrated in the following two examples.

Example 2.7

Find the derivative of $y = x^x$.

Solution: For this example assume $x > 0$ (since $x$ is both the base and the exponent). Note that you cannot use the Power Rule for this function since the exponent $x$ is a variable, not a fixed number. Instead, take the natural logarithm of both sides of the equation $y = x^x$ and then take the derivative of both sides and solve for $y'$:

\begin{align*} \ln\,y ~&=~ \ln\,\left(x^x\right) ~=~ x \;\cdot\; \ln\,x\\ \ddx\,(\ln\,y) ~&=~ \ddx\,(x \;\cdot\; \ln\,x)\\[4pt] \frac{y'}{y} ~&=~ 1 \;\cdot\; \ln\,x ~+~ x \;\cdot\; \frac{1}{x}\\[4pt] y' ~&=~ y \,(\ln\,x ~+~ 1) ~=~ x^x\,(\ln\,x ~+~ 1) \end{align*}

Example 2.8

Find the derivative of $y = \frac{(2x + 1)^7 (3x^3 - 7x + 6)^4}{(1 + \sin x)^5} $.

Solution: Use logarithmic differentiation by taking the natural logarithm of $y$ and then use properties of logarithms to simplify the differentiation before solving for $y'$:

\begin{align*} \ln\,y ~&=~ \ln\,\left(\frac{(2x + 1)^7 (3x^3 - 7x + 6)^4}{(1 + \sin\,x)^5}\right) ~=~ \ln\,\left((2x + 1)^7 (3x^3 - 7x + 6)^4\right) ~-~ \ln\,\left((1 + \sin\,x)^5\right)\\[6pt] &=~ \ln\,\left((2x + 1)^7\right) ~+~ \ln\,\left((3x^3 - 7x + 6)^4\right) ~-~ \ln\,\left((1 + \sin\,x)^5\right)\\[4pt] \ddx\,(\ln\,y) &=~ \ddx\,\left(7\,\ln\,(2x + 1) ~+~ 4\,\ln\,(3x^3 - 7x + 6) ~-~ 5\,\ln\,(1 + \sin\,x)\right)\\ \frac{y'}{y} ~&=~ 7 \cdot \frac{2}{2x + 1} ~+~ 4 \cdot \frac{9x^2 - 7}{3x^3 - 7x + 6} ~-~ 5 \cdot \frac{\cos\,x}{1 + \sin\,x}\\[4pt] y' ~&=~ y \cdot \left(\frac{14}{2x + 1} ~+~ \frac{36x^2 - 28}{3x^3 - 7x + 6} ~-~ \frac{5 \cos\,x}{1 + \sin\,x}\right)\\[6pt] &=~ \frac{(2x + 1)^7 (3x^3 - 7x + 6)^4}{(1 + \sin x)^5} \cdot \left(\frac{14}{2x + 1} ~+~ \frac{36x^2 - 28}{3x^3 - 7x + 6} ~-~ \frac{5 \cos\,x}{1 + \sin\,x}\right) \end{align*}

Radioactive Decay

A classic example of the differential equation $\dydt = ky$ is the case of exponential decay of a radioactive substance, often referred to simply as radioactive decay. In this case the general solution $y = Ae^{kt}$ represents the amount of the substance at time $t \ge 0$, and the decay constant $k$ is negative: $\dydt < 0$ since the substance is decaying (i.e. the amount of substance is decreasing) while $y > 0$ , so $\dydt = ky$ implies that $k < 0$.

tikzpicture[>=latex, every node/.style=font=,domain=0:5.6] [dashed] (0,1.5) -- (1.654, 1.5) -- (1.654,0); [<->,black!60,line width=1pt] (0,4) node[above] y |- (5.8,0) node[right] t; [below] at (0,0) 0; [left] at (0,3) A_0; [left] at (0,1.5) 12 A_0; [above right] at (2,2) y = A_0 e^kt (k < 0); [linecolor,line width=1.5pt] plot ( ,3*exp(-0.419* )); [below] at (1.654,0) t_H; (0,3) circle (2.5pt); tikzpicture — Figure 2.3.3

The constant $A$ is the initial amount of the substance, i.e. the amount at time $t = 0$: $y(0) = Ae^{0t} = Ae^0 = A$. For this reason $A$ is sometimes denoted by $A_0$. The constant $k$ turns out to be related to the half-life of the substance, defined as the time $t_H$ required for half the current amount of substance to decay (see Figure 2.3.3).

You might be tempted to think that the half-life is not a constant, that it might change depending on the amount of substance present. For example, perhaps it would take longer for 100 g of the substance to decay to 50 g than it would for 10 g to decay to 5 g. However, this is not so. To see why, pick any $t \ge 0$ as the current time, so that $y(t) = A_0 e^{kt}$ is the current amount of the substance. By definition, that amount should be halved when the time $t_H$ has passed, that is, $y(t + t_H) = \frac{1}{2} y(t)$. Then $t_H$ does turn out to be independent of the initial amount $A_0$ and depends only on $k$, since

\begin{align*} y(t + t_H) ~=~ \frac{1}{2} y(t) \quad&\Rightarrow\quad A_0 e^{k(t + t_H)} ~=~ \frac{1}{2} A_0 e^{kt} \quad\Rightarrow\quad \cancel{A_0} \cancel{e^{kt}} \cdot e^{kt_H} ~=~ \frac{1}{2}\cancel{A_0} \cancel{e^{kt}}\\[4pt] &\Rightarrow\quad e^{kt_H} ~=~ \frac{1}{2} \quad\Rightarrow\quad kt_H ~=~ \ln\,\left(\frac{1}{2}\right) ~=~ -\ln\,2 \end{align*}

and so:

\begin{equation} t_H ~=~ -\frac{\ln\,2}{k} \qquad\text{and}\qquad k ~=~ -\frac{\ln\,2}{t_H} \tag{2.6}\end{equation}

Example 2.9

Suppose that 5 mg of a radioactive substance decays to 3 g in 6 hours. Find the half-life of the substance.

Solution: Consider $A_0 = 5$ mg as the initial amount, so that $y(t) = 5 e^{kt}$ is the amount at time $t \ge 0$ hours. Use the given information that $y(6) = 3$ mg to find $k$, the decay constant of the substance:

\[3 ~=~ y(6) ~=~ 5 e^{k6} \quad\Rightarrow\quad 6k ~=~ \ln\,\left(\frac{3}{5}\right) \quad\Rightarrow\quad k ~=~ \frac{1}{6}\,\ln\,0.6\]

Then the half-life $t_H$ is:

\[t_H ~=~ -\frac{\ln\,2}{k} ~=~ -\frac{\ln\,2}{\frac{1}{6}\,\ln\,0.6} ~=~ 8.14 ~\text{hours}\]

Note in the above example that the given time $t = 6$ was used for finding the constant $k$ and then the half-life $t_H$. For the converse problem—given the half-life find the time required for a certain amount to decay—you would do the opposite: use the given $t_H$ to find $k$ and then solve for the required time $t$ from the equation $y(t) = A_0 e^{kt}$.

Example 2.10

Another example of the differential equation $\dydt = ky$ is exponential growth of cell bacteria, in which case $k > 0$ since the number of cells $y(t)$ at time $t$ is increasing.

Example 2.11

Another example is for the current $I$ in a simple series electric circuit with a constant direct current (DC) source of voltage $V$, a capacitor with capacitance $C$, a resistor with resistance $R$, and a switch, as in Figure 2.3.4. If the capacitor is initially uncharged when the switch is open, and if the switch is closed at time $t = 0$, then the current $I(t)$ through the circuit at time $t \ge 0$ satisfies (by Kirchoff’s Second Law) the differential equation

\[RC\frac{d\negmedspace I}{\dt} ~+~ I ~=~ 0 \quad\Rightarrow\quad \frac{d\negmedspace I}{\dt} ~=~ -\frac{I}{RC}\]

so that $I(t) = I_0 e^{-t/RC}$ where $I_0$ is the initial current at $t = 0$. Ohm’s Law says that $V = I_0R$, so

\[I(t) ~=~ \frac{V}{R} e^{-t/RC}\]

is the current at time $t \ge 0$, which decreases exponentially.

Example 2.12

In the previous examples the quantities that decayed or grew exponentially did so as functions of time. There are other possible variables besides time, though. For example, the atmospheric pressure $p$ measured as a function of height $h$ above the surface of the Earth satisfies—assuming constant temperature—the differential equation

\[\frac{d\negmedspace p}{d\!h} ~=~ -\frac{w_0}{p_0}\,p\]

where $p_0$ is the pressure at height $h = 0$ (i.e. ground level) and $w_0$ is the weight of a cubic foot of air at pressure $p_0$ (with air pressure measured in lbs per square foot and height measured in feet). Thus,

\[p(h) ~=~ p_0\;e^{-\frac{w_0}{p_0}\,h} ~.\]

So the atmospheric pressure decreases exponentially as the height above the ground increases.

Exercises

For Exercises 1-12, find the derivative of the given function.

$y ~=~ e^{2x}$
$y ~=~ xe^{x^2}$
$y ~=~ e^{-x} ~-~ e^{x}$

3
$y ~=~ e^{\sin\;x}\vphantom{\dfrac{1 ~+~ e^x}{1 ~-~ e^x}}$
$y ~=~ \dfrac{1 ~+~ e^x}{1 ~-~ e^x}$
$y ~=~ \dfrac{1}{1 ~+~ e^{-2x}}$

3
$y ~=~ e^{e^x}$
$y ~=~ e^{2\,\ln\,x}$
$y ~=~ \ln\,(3x)$

3
$y ~=~ \ln\,(x^2 ~+~ 2x ~+~ 1)^4$
$y ~=~ \left(\ln (\tan\;x^2 )\right)^3$
$y ~=~ \ln\,(e^x ~+~ e^{2x})$
Show that $~\dfrac{d}{\dx}\,\left(\ln\,(kx)\right) ~=~ \dfrac{1}{x}~$ for all constants $k > 0$.
Show that $~\dfrac{d}{\dx}\,\left(\ln\,\left(x^n\right)\right) ~=~ \dfrac{n}{x}~$ for all integers $n \ge 1$.

For Exercises 15-18, use logarithmic differentiation to find $\dydx$.

$y ~=~ x^{x^2}\phantom{\dfrac{x^2}{x^3}}\vphantom{\left(\dfrac{x^3}{2}\right)^{8}}$
$y ~=~ x^{\ln\,x}\phantom{\dfrac{x^2}{x^3}}\vphantom{\left(\dfrac{x^3}{2}\right)^{8}}$
$y ~=~ x^{\sin\,x}\phantom{\dfrac{x^2}{x^3}}\vphantom{\left(\dfrac{x^3}{2}\right)^{8}}$
$y ~=~ \dfrac{(x+2)^{8} \, (3x-1)^{7}}{(1-5x)^4}\vphantom{\left(\dfrac{x^3}{2}\right)^{8}}$

Suppose it takes 8 hours for 30% of a radioactive substance to decay. Find the half-life of the substance.
The radioactive isotope radium-223 has a half-life of 11.43 days. How long would it take for 3 kg of radium-223 to decay to 1 kg?
If a certain cell population grows exponentially—i.e. is of the form $A_0e^{kt}$ with $k>0$—and if the population doubles in 6 hours, how long would it take for the population to quadruple?

For Exercises 22-25, use induction to prove the given formula for all $n \ge 0$.

$\dfrac{d^n}{\dx^n}\,\left(e^{kx}\right) ~=~ k^n \,e^{kx}\quad$ (any constant $k \ne 0$)
$\dfrac{d^n}{\dx^n}\,\left(x\,e^x\right) ~=~ (x + n)\,e^x$

2
$\dfrac{d^n}{\dx^n}\,\left(x\,e^{-x}\right) ~=~ (-1)^n (x - n)\,e^{-x}\vphantom{\dfrac{d^{n+1}}{\dx^{n+1}}}$
$\dfrac{d^{n+1}}{\dx^{n+1}}\,\left(x^n\;\ln\,x\right) ~=~ \dfrac{n!}{x}$
Show that $f'(x) = f(x)\,(1 - f(x))$ for the sigmoid neuron function $f(x) = \dfrac{1}{1 + e^{-x}}$. This derivative relation is used in neural network learning algorithms.
If $\;y = C e^{-\kappa t}\,\cos\,\left(\sqrt{n^2 - \kappa^2}\;t ~+~ \gamma\right)~$ then show that

\[\frac{d^2y}{\dt^2} ~+~ 2\kappa\,\dydt ~+~ n^2 y ~=~ 0\]

for all constants $C$, $n$, $\kappa$, $\gamma$, with $0 \le \kappa \le n$.

Suppose that $~e^y + e^x ~=~ e^{y + x}$. Show that $\dydx = -e^{y-x}$.
For an infinitesimal $\dx$ show that $e^{\dx} = 1 \;+\; \dx$. (Hint: Use $\ddx\,(e^x) = e^x$.)
For an infinitesimal $\dx$ show that $\ln\,(1 + \dx) = \dx$.

For the general case see pp.61-63 in Franklin, P., A Treatise on Advanced Calculus, New York: Dover Publications, Inc., 1964. ↩
For example, to see how square roots and cube roots are calculated see Chapter 2 in Fike, C.T., Computer Evaluation of Mathematical Functions, Englewood Cliffs, New Jersey: Prentice-Hall, Inc., 1968. ↩
It will be shown in Chapter 9 that this limit does in fact exist. ↩
This is admittedly a “hand waving” argument. In Chapter 3 a more exact method will be discussed for proving limits like this one. ↩
This text almost used $\log\,x$ as well, prevented only by the desire for compatibility with other mathematics texts. ↩
Note that when using the formula $\ln\,\left(\frac{a}{b}\right) = \ln\,a - \ln\,b$ in numerical computations—especially on hand-held calculators—it is preferable to use the left side of the equation, i.e. $\ln\,\left(\frac{a}{b}\right)$, since the right side $\ln\,a - \ln\,b$ is vulnerable to the problem of subtractive cancellation, which can give an incorrect answer of 0 if $a$ and $b$ are nearly equal. For a discussion of subtractive cancellation see § 1.3 in Henrici, P., Essentials of Numerical Analysis, with Pocket Calculator Demonstrations, New York: John Wiley & Sons, Inc., 1982. ↩