Chapter 7

Exponential Functions

and Logarithms

Chess and Rice

Enamored of chess, the emperor summoned the game’s creator to his palace and offered him anything he wished. Much to the emperor’s surprise, this ingenious gentleman asked only for rice: two grains for the chessboard’s first square, four for its next square, then eight, sixteen, and so forth. The emperor nodded assent, and called a slave to do the counting.

As the emperor and the man spoke of this and that, the slave’s rice bag ran out. He fetched a second bag, but this too was soon emptied. Undaunted, he brought several more, but these were exhausted in turn. Again and again the slave fetched more rice, until the emperor, observing that his slave had still progressed no further than the chessboard’s second row, threatened him terribly. In the ensuing silence, knowledge dawned in emperor and slave: The task was impossible – were all the rice in the world gathered together at the palace, it would still prove insufficient to the task. The emperor, horrified, tore his robes and left the palace, never to be seen again. The slave hanged himself in the garden. The man returned home.

• Long Shu, Annals, Vol. 3, Part 11.

In the preceding account, the man requests 2 grains for the chessboard’s first square, $ 2^{2} $ for its second square, $ 2^{3} $ grains for its third square and so forth. Stated algebraically, the board’s $ n^{th} $ square will have $ 2^{n} $ grains. As the emperor and slave come to realize, the function $ y = 2^{n} $ grows with alarming speed. You can work out the number of grains for the first ten or so squares in your head, but soon, you’ll want a calculator. A calculator for example, tells us that the $ 32^{nd} $ square should have 4,294,967,296 grains. Since most calculators give exact values only for numbers of ten or fewer digits, you’ll need a computer (or sufficient paper, ink, and patience) to discover that the chessboard’s $ 64^{th} $ (and last) square will have precisely 18,446,744,073,709,551,616 grains of rice – considerably more rice than exists in the world.

In this chapter, we’ll consider functions such as $ f(x) = 2^{x} $ , which are defined for all real numbers (not just for whole numbers). Such a function, whose independent variable appears as an exponent, is called an exponential function.

Exercises

The rice story explains why $ f(x) = 2^{x} $ grows so rapidly, but it doesn’t explain all the features of its graph. To understand why its graph (shown at right) looks as it does…

1. Explain why the graph’s y-intercept is 1.

2. We know that $ f(1) = 2 $ , and $ f(2) = 4 $ . What is $ f(1.5) $ ? (Give exact and approximate answers.)

3. Is the negative x-axis an asymptote of the graph? How do you know?

Exponential Growth

The greatest shortcoming of the human race is our inability to understand the exponential function.

• Albert Bartlett

Exponential functions of the form $ y = a^{x} $ (where a > 1) are said to exhibit exponential growth. The greater the base, a, the steeper the growth, as we see by comparing the graphs at right. Exponential growth occurs whenever something increases by a fixed percentage per unit of time, as I’ll soon explain.

A common example is compound interest at a bank. Of course, your money in the bank is on the early, slowly growing part of the curve, and it will be a long time before its growth becomes explosive. By that time, you won’t be around to enjoy it.

\[ y=5^{x} \]

\[ y=2^{x} \]

A potentially dangerous example of exponential growth is Earth’s human population. Consider yourself lucky if you aren’t around to witness the consequences of that exponential explosion.

Example. Suppose 100 bacteria are in an enclosed space. Every minute, the bacteria population increases its size by 64%. How many bacteria will there be after 1 minute? After 2 minutes? After 3? Express the number of bacteria as a function of the number of minutes that have elapsed, and graph this function.

Solution. The population increases by 64% each minute, so in the first minute, the population grows from 100 to 100(1.64). In the second minute, it grows from 100(1.64) to 100(1.64) $ ^{2} $ . In the third, it grows from 100(1.64) $ ^{2} $ to 100(1.64) $ ^{3} $ . Following this pattern, we see that after t minutes, the population will reach 100(1.64) $ ^{t} $ . Thus, if $ B(t) $ represents the number of bacteria at time t, then

\[ B(t)=100(1.64)^{t}. \]

This is easy to graph by thinking about transformations. First, we know that the graph of the simpler function $ B = (1.64)^{t} $ looks like the standard picture of exponential growth described at the top of this page: It creeps up from the negative x-axis, cuts through the y-axis at $ (0,1) $ , and takes off into the stratosphere. Stretching this graph vertically by a factor of 100 will turn it into the graph we seek. The stretch will not particularly change the overall shape of the graph (as you should convince yourself) except that now its y-intercept will be 100 instead of 1. It will therefore look like the figure at right.

Exercises

4. In the Example above, find the number of bacteria (rounded to the nearest bacterium) in the enclosed space after 1 minute, 2 minutes, 3 minutes, 30 minutes, and an hour.

5. Suppose a piece of paper is $ 1/4 , mm $ thick. If you cut it in half and stack the halves, the result will be $ 1/2 , mm $ thick. Repeat the operation on the new stack, and the result will be $ 1 , mm $ thick. By playing around with a calculator, determine…

1. the number of times you must cut and restack to form a pile of paper taller than you.

2. the number of times you must you cut and restack to form a pile of paper that will reach the moon.

[Hint: The moon is, on average, 384,400 km from the earth.]

6. Suppose that a certain bank account pays 2% interest at the end of each year. If you deposit $2000 into this account, and then turn your back on it as it accumulates interest, how much money will be in the account after 1 year? After 2 years? After 20 years? After 200 years? (Give your answers to the nearest dollar.)

7. Sketch graphs of the following exponential functions, noting asymptotes and intercepts:

\[ y=3^{x} \]

2. y = 42(10x)

\[ f(x)=20(5^{x}) \]

\[ g(x)=3(2^{x})+5 \]

5. $ y = 12(8^{x}) - 10 $ f) $ y = 2^{-x} $ [An example of exponential decay, which we’ll soon discuss.]

8. As it is with bacteria, so it is with people.

The world’s human population is growing exponentially, and our species has clearly gone quite around the bend already. At present, the world’s human population is increasing at a rate of about 1% per year. Should this growth rate remain constant, then, given that the world’s population in 2022 was 8 billion, what will it be (to the nearest tenth of a billion) in the year 2050? In 2075? In 2100?

9. In the year 1990, the population of Arse, Wisconsin was 950. Each subsequent year, the population has increased by 10 people. Assuming this trend continues, express the population of Arse as a function of t, the number of years that have elapsed since the year 2000. What is the projected population of Arse in 2050?

10. In the previous problem, change the word “people” to “percent”, and redo the problem. Then ponder the distinction between linear growth and exponential growth.

11. Some exponent review. Simplify the following as much as possible:

\[ \begin{array}{ccc}a)\frac{\left(5^{-1}5^{3}\right)^{2}}{5^{4}}&\quad&b)8^{2/3}\quad&c)\left(\frac{\left(6^{7}6^{3}\right)^{2}}{(6^{6})^{3}}\right)^{1.5}\quad&d)\left(2^{5}(4^{-2})\big(16^{3/4}\big)\right)^{-3/2}\quad&e)\left(\frac{3^{\pi}e^{2/3}e^{1/6}}{3^{e}e^{\pi}}\right)^{0}\end{array} \]

Exponential Decay

Whenever a population increases by a fixed percentage per unit time, the result is exponential growth. Whenever a population decreases by a fixed percentage per unit time, the result is exponential decay.

Example 1. The population of Storyproblem, Ohio peaked in the year 1980, when it boasted 30,000 citizens, most of whom were gainfully employed measuring flagpoles, computing the ideal dimensions of fenced enclosures, and so forth. Since its peak, however, the town’s population has declined by 3% each year. Express the population of Storyproblem as a function of t, the number of years that have passed since 1980, and find the population in the year 2020.

Solution. After 1 year, the population was 97% of what it had been. That is, it was 30,000(.97). After 2 years, the population was only 97% of this. That is, the population fell to 30,000(.97) $ ^{2} $ . And clearly, after t years, the population P of Storyproblem will be

\[ \mathbf{\boldsymbol{P}}(\mathbf{\boldsymbol{t}})=3\mathbf{\boldsymbol{0}},\mathbf{\boldsymbol{0}}\mathbf{\boldsymbol{0}}\mathbf{\boldsymbol{0}}(\mathbf{\boldsymbol{.}}\mathbf{\boldsymbol{97}})^{\mathbf{\boldsymbol{t}}}. \]

In particular, in 2020 (i.e. 40 years after 1980), the town’s population was $ P(40) $ .

The heart of any exponentially decaying function is an equation that has the form $ y = a^{x} $ , where $ a < 1^{*} $ . (For example, the heart of the function in the example above is $ y = (.97)^{x} $ .) Be sure you understand why this is so. The graphs of such functions turn out to be the mirror images of their exponentially growing siblings, which is easy to see with the help of some algebraic trickery.

Consider, for example, the function $ y = (.8)^{x} $ , which describes a population declining by 20% with each unit of time that passes. We can rewrite our function as

\[ y=(.8)^{x}=\left(\frac{8}{10}\right)^{x}=\left(\frac{10}{8}\right)^{-x}=(1.25)^{-x}. \]

Provided you remember the previous chapter’s material on transformations, you should recognize that this function’s graph is simply a reflection of $ y = (1.25)^{x} $ (an exponential growth curve) over the y-axis. A similar analysis can be carried out for any exponentially decaying function. Thus all exponentially decaying functions of the form $ y = a^{x} $ exhibit the same behavior: They descend from the heavens, cross the y-axis at $ (0,1) $ , and slink off asymptotically towards zero.

Perhaps the most astonishing application of exponential decay is radiocarbon dating, about which I’ll have more to say later in the chapter.

Exercises

12. The local bordello raises its price of admission by 10%, but then, one week later, decides to lower the price by 10%. What is the net effect of these two changes?

13. At the age of 35, a man has 1000 marbles. Each year thereafter, he loses 1.9% of his marbles. Express the number of marbles in his possession as a function of t, the number of years that have passed since his $ 35^{th} $ birthday. [These are special mathematical marbles, somewhat liquid in nature; one can have a fractional number of them.]

14. If we could reverse time, and watch the man from the previous problem growing ever younger, and thus gaining new marbles each year, would his marble collection grow by 1.9% each year?

15. Sketch graphs of the following exponential functions, noting asymptotes and intercepts:

\[ y=(.6)^{x} \]

\[ \mathsf{b})y=232(.6184)^{x} \]

\[ f(t)=\pi\left(\frac{3}{14}\right)^{x} \]

\[ \mathsf{d})g(x)=3(2^{-x})+5 \]

\[ y=-12(8^{x})-10 \]

\[ \mathsf{f})y=(2.71828)^{x} \]

\[ y=-\frac{1}{2}\Big(\frac{1}{8}\Big)^{x}+1 \]

16. Express each of the following functions in the form $ y = ba^{t} $ , where 0 < a < 1, and state the percentage of the function’s value that is lost per unit of time (t).

1. $ y = 5^{-t} $

2. $ y = 2(2)^{-t} $

3. $ y = ()^{-t} $

17. Express each of the following functions in the form $ y = ba^{-t} $ , where a > 1, and state the percentage of the function’s value that is lost per unit of time (t).

\[ y=(.61)^{t} \]

\[ y=4\left(\frac{1}{10}\right)^{t} \]

\[ y=\frac{7}{8}\left(\frac{3}{5}\right)^{t} \]

18. Watch at least the first 20 minutes of Albert Bartlett’s lecture “Arithmetic, Population, and Energy” online. After you’ve learned about logarithms, you’ll be able to understand the mathematical basis of the material that he describes: the fixed doubling times of exponential functions and the “rule of 70”. I’ll explain both of these at the end of this chapter, but Bartlett will provide some provocative context.

Logarithms: a Tool for Solving Exponential Equations

Logarithm – had I – for Drink –

’Twas a dry Wine –

• Emily Dickinson (“Let Us play Yesterday”)

Next, we’ll develop a tool – the natural logarithm – for solving exponential equations (such as $ 2^{x} = 5 $ ). This tool involves an irrational number, e, whose nature must, alas, remain mysterious until you study derivatives in calculus. Trying to understand why e is “natural” without understanding derivatives is like trying to understand $ $ without understanding circles. If you don’t believe me, just read an account of e in any other precalculus textbook. You’ll probably be treated to a contrived explanation involving “continuously compounded” interest that you’ll not find the least bit enlightening. Pay no mind. In this course, all you must know about e is that it is a number and that $ e $ .

The Natural Logarithm Defined

The graph of $ y = e^{x} $ at right is a typical picture of exponential growth. If you can answer the following simple questions about it, you already understand the natural logarithm.

To what power must we raise e to get 1?

(Answer: 0, since $ e^{0} = 1 $ .)

To what power must we raise e to get 2?

(Answer: about 0.7, which we see from the graph.)

To what power must we raise e to get 0.5?

(Answer: about -0.7, which we see from the graph.)

To what power must we raise e to get $ $ ?

(Answer: about 1.2, which we can see from the graph, since $ $ )

To what power must we raise e to get e?

(Answer: exactly 1, since $ e^{1} = e $ .)

Without realizing it, you’ve just been answering questions about the natural logarithm.

Definition. The natural logarithm of x is the power to which we must raise e to get x.

Writing the phrase “the natural logarithm of x” grows tiresome, so we adopt a shorthand notation for it: $ x $ . Using this notation, we can rewrite the questions and answers above in the following compact form: $ (1)=0 $ ; $ (2) $ ; $ (0.5) $ ; $ () $ ; $ (e)=1 $ .

The natural logarithm is sufficiently important to merit its own button on scientific calculators. Our visual estimates of $ (2) $ , $ (0.5) $ and $ () $ in the last paragraph are crude, but with a calculator, we can obtain highly accurate approximations – accurate to as many decimal places as your calculator will display. Running these three logarithms through mine, I find that

\[ \ln(2)\approx0.693147181;\qquad\ln(0.5)\approx-0.693147181;\qquad and\qquad\ln(\pi)\approx1.144729886. \]

These are accurate to the nearest billionth, which is more than ample for any scientific application. Notice the surprising relationship between $ (2) $ and $ (0.5) $ ; they are negatives of one another! This is not just a crazy coincidence. You will understand why this is true after you’ve learned some of the most important properties of the logarithm.

The Natural Logarithm’s Inverse Properties

What color was George Washington’s white horse?

• Groucho Marx

By definition, $ (e^{x}) $ is the power to which we must raise e to get $ e^{x} $ . That power is obviously x, so

\[ \mathbf{ln}(e^{x})=x. \]

This is the first of the natural logarithm’s two vital “inverse properties.” If you understand the natural logarithm’s definition, the inverse properties are as obvious as the answer to Groucho’s question.

Obviously, if we raise e to the-power-to-which-we-must-raise-e-to-get-x, then the result will be… x. Translating this obvious statement into symbols, we have

\[ e^{\ln\left(x\right)}=x. \]

This is the second inverse property of logarithms. (Oh, and by the way, who is buried in Grant’s tomb?) Bear in mind that in each of these identities, x stands for anything whatsoever. So for example, we have

\[ \ln(e^{3t+7})=3t+7,\quad\ln(e^{666})=666,\quad\ln\left(e^{\textcircled{\textup{c}}}\right)=\textcircled{\textup{c}} \]

\[ e^{\ln(999)}=999,\qquad e^{\ln(72x^{2}+14)}=72x^{2}+14,\qquad e^{\ln(\textcircled{\otimes})}=\textcircled{\otimes}. \]

The moral of the story is that the functions $ e^{x} $ and $ (x) $ “undo” one another: If you apply one of them to an expression, and then immediately apply the other, you will go right back to the original expression. This “undoing” business is rather like adding two and then subtracting two, or taking a cube root and then cubing the result. Functions that “undo” one another are called inverse functions, which is why I’ve called these two properties inverse properties.

Exercises

19. Simplify: a) $ (e^{32}) $ b) $ (e^{5x}) $ c) $ (e) $ d) $ e^{} $ e) $ e^{(x{2}+1)} $ f) $ e^{(e{})} $

20. (An introduction to inverse functions in general)

We define $ f^{-1}(x) $ (which we pronounce “f inverse of x”) to be the number that f sends to x.

The function $ f^{-1} $ defined thereby is called $ f’ $ inverse function.

Here are some problems for you to help you get a feel for this notion (and notation):

1. If $ g(x) = 4x $ , find $ g^{-1}(8) $ .

2. If $ h(x) = 4x^{3} + 2 $ , find $ h^{-1}(-2) $ .

3. If $ k(x) = e^{x} $ , find $ k^{-1}(2) $ .

4. Inverse functions “undo” one another; applying them successively to an expression yields the same expression with which you started. Using the definition above, explain why $ f(f^{-1}()) = $ .

[Hint: What city is famous for Boston Baked Beans?]

5. This abstract idea of an inverse function is vital in higher mathematics, but not in (pre)calculus, where the few inverses you’ll use are concrete and have their own specific notation, rendering the $ f^{-1} $ business superfluous. [For example, the number that $ f(x) = e^{x} $ sends to x we call $ x $ , not $ f^{-1}(x) $ .] This is just as well, because the abstract notation for an inverse function was poorly designed (alas, it’s far too late to change it now), and confuses many precalculus students. Note well: Although the $ -1 $ in $ f^{-1}(x) $ looks like an exponent, it isn’t! Your problem: if $ g(x) = $ , find $ g^{-1}(10) $ and $ [g(10)]^{-1} $ . Verify that these are unequal.

The Natural Logarithm’s Exponent Property

When solving exponential equations, our main tool will be the following property of logarithms:

\[ \mathbf{ln}\big(\boldsymbol{a}^{b}\big)=\boldsymbol{b}\mathbf{ln}(\boldsymbol{a}). \]

Observe what happens here: The logarithm pulls the exponent down and then turns it into a mere factor, which will let us reduce a problem involving exponentiation to a problem involving mere multiplication. We can prove this “exponent property of logarithms” as follows.

Claim. For any a and b for which the following expressions are defined, $ (a^{b}) = b (a) $ .

Proof. Because $ a = e^{(a)} $ (by an inverse property of logarithms), we can substitute $ e^{(a)} $ for a. Doing this in the expression $ (a^{b}) $ , we obtain

\[ \begin{aligned}\ln\big(a^{b}\big)&=\ln\Big(\big(e^{\ln(a)}\big)^{b}\Big)\quad&(by an inverse property of logarithms)\\ &=\ln\big(e^{b\ln(a)}\big)\quad&(by a familiar property of exponents)\\ &=b\ln(a)\quad&(by the other inverse property of logarithms).\end{aligned} \]

Thus we’ve proved that $ (a^{b})=b(a) $ , as claimed.

That first move in the preceding proof (writing a as $ e^{(a)} $ ) comes in handy elsewhere in mathematics. It’s just another variation on the algebraic theme of changing a thing’s form while preserving its value. Note the elegant logical flow of the last few sections: The natural logarithm’s definition led immediately to the logarithm’s inverse properties, which we then used to obtain the logarithm’s exponent property.

We’re now able to understand why $ (0.5) = -(2) $ , a curious fact that we stumbled upon a few pages ago when we were playing with our calculators. Here’s the explanation:

\[ \ln(0.5)=\ln\left(\frac{1}{2}\right)=\ln(2^{-1})=(-1)\ln(2)=-\ln(2).^{*} \]

Finally, I want to alert you to a common mistake. All too often, beginners mistakenly suppose that, for example, $ (ab^{c}) $ equals $ c (ab) $ . The exponent property of logarithms does not apply in this case, because the exponent c covers only b, not ab. [In contrast, it is quite true that $ ((ab)^{c}) = c (ab). $$ The following exercise should help you understand this small but important point.

Exercise.

21. True or False:

\[ a)\ln(x^{3})=3\ln x \]

\[ b)2\ln3=\ln9 \]

\[ \mathsf{c})\ln\bigl(2x^{3}\bigr)=3\ln(2x) \]

\[ \mathsf{d})\ln\left(8x^{3}\right)=3\ln(2x) \]

5. $ 3 (2a) = 8a^{3} $

\[ \mathsf{f})e\ln(e^{3})=3e \]

\[ \mathfrak{g})\ln(a^{2}b^{2})=2\ln(ab) \]

\[ h)\ln\left(ab^{5}\right)=5\ln(ab) \]

Solving Exponential Equations with Logarithms

…as an insane man mistakes his visiting kin for galaxies, logarithms, low-haunched hyenas – but there are also madmen – and they are invulnerable – who take themselves for madmen – and here the circle closes.

• Cincinnatius C, from Vladimir Nabokov’s Invitation to a Beheading (Chapter 13).

By taking the natural logarithm of both sides of an exponential equation, we can pull the variable down to the ground where we can at it.

Example 1. Solve the equation $ 2^{x} = 5 $ .

Solution.

\[ \begin{aligned}2^{x}&=5\\\ln(2^{x})&=\ln\left(5\right)&\quad\left(taking the natural logarithm of both sides^{*}\right)\\x\ln(2)&=\ln\left(5\right)&\quad\left(by the exponential property of the natural logarithm\right)\\x&=\frac{\ln\left(5\right)}{\ln\left(2\right)}.\end{aligned} \]

Thus the equation’s exact solution is $ (5)/(2) $ . Should you need a decimal approximation for it, you can of course get one from your calculator, which will tell you that $ x $ .

In almost every precalculus class, some student will imagine that in an expression such as $ (5)/(2) $ , he or she can “cancel the $ ’s $ ”, thereby “simplifying” it to 5/2. Please do not be this person. Do not forget: You may only cancel common factors from the numerator and denominator of a fraction.

Example 2. Solve the following equation: $ 2 = 5(7)^{-2t} $

Solution. That factor of 5 would cause us some trouble were we to take natural logarithms of both sides right away, so to avoid that, we’ll begin by dividing both sides by 5. The rest is straightforward:

\[ \begin{aligned}2&=5(7)^{-2t}\\2/5&=7^{-2t}\\\ln(2/5)&=\ln(7^{-2t})\\\ln(2/5)&=(-2t)\ln(7)\\t&=\frac{\ln\left(2/5\right)}{-2\ln(7)}\approx0.235\end{aligned} \]

(by the exponential property of ln)

Exercise.

22. Solve for the given variable. (Give exact and approximate answers.)

\[ a)5^{x}=14 \]

\[ \mathsf{b})3^{x}=1/4 \]

\[ e^{5x}=17 \]

\[ \mathsf{d})4^{4x-5}=38 \]

\[ \mathrm{e)}\ 1000(1.03)^{t}=5000 \]

\[ 18(1.06)^{t}=550 \]

\[ g)50e^{-0.12t}=10 \]

\[ 100-100\left(\frac{1}{4}\right)^{x}=70 \]

\[ 13+8(10^{x})=20 \]

\[ \mathrm{j})\;5e^{0.02t}=3 \]

The Natural Logarithm’s Multiplication and Division Properties

The exponent property of logarithms turns exponentiation into the simpler operation of multiplication. Similarly, the following “multiplication property” of logarithms reduces multiplication to addition:

Claim 1. For any a and b for which the following expressions are defined, we have

\[ \mathbf{ln}(ab)=\mathbf{ln}(a)+\mathbf{ln}(b). \]

Proof. If you compare them closely, you’ll see that this proof is basically the same as our proof of the exponent property. Be sure that you understand each step.

\[ \begin{aligned}\ln(ab)&=\ln\big(e^{\ln(a)}e^{\ln(b)}\big)\quad&(by an inverse property of\ln)\\&=\ln\big(e^{\ln(a)+\ln(b)}\big)\quad&(by the ordinary algebra of exponents)\\&=\ln(a)+\ln(b)\quad&(by the other inverse property of\ln)\end{aligned} \]

Thus we’ve proved that $ (ab)=(a)+(b) $ , as claimed.

There’s an analogous “division property” of logarithms that reduces division to subtraction.

Claim 2. For any a and b for which the following expressions are defined,

\[ \mathbf{l n}\left(\frac{a}{b}\right)=\mathbf{l n}(a)-\mathbf{l n}(b). \]

Proof. (I leave this as an exercise for you. If you understood the proofs of the exponent property and the multiplication property, this should be fairly easy.)

You probably won’t use the multiplication and division properties of logarithms very often in subsequent courses, but they certainly do come up from time to time, so you should know them. It’s traditional for precalculus books to include reams of problems such as the following:

Problem. Simplify the expression $ (3x) + (5x^{2}) - (3) $ by writing it as a single logarithm.

\[ \begin{aligned}Solution.\ \ln(3x)+\ln(5x^{2})-\ln(3)&=\ln(3x\cdot5x^{2})-\ln(3)\quad(by the multiplication property)\\&=\ln(15x^{3})-\ln\ (3)\\&=\ln(15x^{3}/3)\quad(by the division property)\\&=\ln(5x^{3}).\end{aligned} \]

Traditional, but not particularly productive. I’ll give you a few too, but I won’t belabor the point.

Exercises

23. Simplify by writing as a single logarithm:

1. $ (6x^{9}) - (3x^{2}) $ b) $ (14t) + (2t^{-1}) $ c) $ 3(2x) - (4x^{2}) $

24. Write as a sum (or difference) of logarithms, each with as simple an argument as possible:

1. $ ( ) $

2. $ ( x^{2}y{3} ) $

3. $ ( x^{3}yx{2}y^{3}x{-5} ) $

The Graph of the Natural Logarithm Function

If you play around with the function $ y = (x) $ , thinking about what it does to various inputs, it’s easy to see that its graph looks like the figure at right. Notice its prominent features:

1. The function is defined only for positive inputs.

[Do you see why? Think about the definition of $ (x) $ .]

2. The negative y-axis is an asymptote of the graph.

[Do you see why? Apply the natural logarithm’s definition to some tiny positive numbers.]

3. As x approaches $ $ , $ (x) $ also approaches $ $ , but it takes its time getting there.

As exponential growth is proverbial for explosively rapid growth, so logarithmic growth is proverbial for slow growth. An exercise below dwells on logarithmic sluggishness.

Graphing $ y = (x) $ and $ y = e^{x} $ on the same set of axes reveals a striking symmetry: The two graphs are mirror images of one another, where the “mirror” is the line y = x.

In fact, if you graph any pair of inverse functions on the same set of axes, they will exhibit the same symmetry about the line y = x, which is occasionally a handy fact to know.

If you understand everything you’ve read up to this point, then you may congratulate yourself: You know just about every important mathematical property of the natural logarithm.

Exercises

25. How large must x be for the value of $ (x) $ to surpass 10? To surpass 100? To surpass 1000?

[Hint: To solve an equation involving $ (x) $ , apply the (base-e) exponential function to both sides.]

26. How do we solve an equation for a variable that lies inside a natural logarithm? We isolate the logarithm on one side of the equation, and then we “undo” the logarithm by applying the base-e exponential function to both sides (as in Example 1 below). Or, if the variable lies inside multiple logarithms (as in Example 2), we first try to combine those logs (by using the multiplication or division properties), and then we proceed as before.

Example 1.

\[ \begin{aligned}1&=3-\ln(2x+1)\\&\ln(2x+1)=2\\&e^{\ln(2x+1)}=e^{2}\\&2x+1=e^{2}\\&x=\frac{e^{2}-1}{2}.\quad\text{♦}\end{aligned} \]

Example 2.

\[ \begin{aligned}\ln x+\ln(2x)&=1\\ \ln(2x^{2})&=1\\ e^{\ln(2x^{2})}&=e^{1}\\ 2x^{2}&=e\\ x&=\sqrt{e/2}\text{．}^{\ast}\quad\text{♦}\end{aligned} \]

Solve the following equations.

1. $ 2 x = 5 $

\[ b)\ln x+\ln\left(\frac{1}{x^{2}}\right)=2 \]

\[ \mathbf{c})\ln(x^{2})-\ln x=1 \]

\[ \mathsf{d})\ln x=\ln(2-x^{2}) \]

“Unnatural” Logarithms: Base-10 (and Others)

Logarithms based on numbers other than e do exist. The base-10 logarithm, for example, often appears in scientific applications, and on rare occasions, base-2 logarithms can be spotted in the wild, usually in the company of computer scientists. Fortunately, everything you know about natural logarithms applies to their unnatural cousins; we need only make some tiny (and obvious) adjustments.

Definition. The base-10 logarithm of x is the power to which we must raise 10 to get x.

Notation. We write $ x $ to denote the base-10 logarithm of x.

Examples. $ = 2 $ (since $ 10^{2} = 100 $ ). $ (0.1) = -1 $ (since $ 10^{-1} = 0.1 $ ).

Scientific calculators have a “log” button for obtaining decimal approximations of base-10 logarithms.

Exercises

27. The inverse properties of the base-10 logarithm are $ (10^{x}) = x $ and $ 10^{(x)} = x $ . Explain in a sentence or two why each of these properties is true. [Hint: Tell ’em Groucho sent you.]

28. The exponent property of the base-10 logarithm is just as you would expect: $ (a^{b}) = b (a) $ .

Prove it. [Hint: Review the corresponding proof for natural logarithms and make appropriate adjustments.]

29. The multiplication and division properties of the base-10 logarithm are also exactly what you expect:

\[ \mathbf{l o g}(a b)=\mathbf{l o g}(a)+\mathbf{l o g}(b),\mathrm{a n d}\mathbf{l o g}(a/b)=\mathbf{l o g}(a)-\mathbf{l o g}(b). \]

Prove these two properties. [Hint: Adapt the hint from the previous exercise.]

30. Sketch graphs of $ y = x $ and $ y = x $ on the same set of axes. [Hint: Think about these logs’ definitions.]

31. Evaluate (without a calculator):

1. $ (1) $ b) $ (1) $ c) $ (e) $ d) $ (10) $ e) $ (10000) $ f) $ () $ g) $ 10^{(42)} $ h) $ (10^{27}) $ i) $ ((10)) $ j) $ ((e)) $ k) $ 10^{((100))} $ l) $ $

32. Solve, giving exact and approximate answers. [Hint (for some parts): To “undo” a base-10 logarithm, apply the base-10 exponential function to both sides of the relevant equation.]

1. $ 2(3x) + 3 = -1 $ b) $ (x^{3}) = 2 $ c) $ 4(2.4)^{x} = 6(1.2)^{x} $

2. $ 5e^{0.12t} = 10e^{0.08t} $ e) $ (x + 5) - (x + 2) = 2 $ f) $ (x) + (x - 3) = 1 $

33. Reflecting the graph of $ y = (x) $ across the line y = x will produce the graph of what function?

34. The base-2 logarithm of x is defined, of course, as the power to which you must raise 2 to get x. It is denoted by the symbol $ {2} x $ . Thus, for example, $ {2} 8 = 3 $ .

1. State the inverse properties of the base-2 logarithm, and explain why they hold.

2. State the exponent property of the base-2 logarithm, and prove it.

3. State the multiplication and division properties of the base-2 logarithm, and prove them.

4. The graph of $ y = _{2} x $ is the reflection across the line y = x of the graph of what function?

Doubling Times

…nor had topsawyer’s rocks by the stream Oconee

exaggerated themselse to Laurens County’s gorgios

while they went doublin their mumper all the time…

– James Joyce, Finnegans Wake

Functions grow at different rates. Some, such as $ y = t^{3} $ , grow rapidly. Others, like $ y = $ , grow slowly. If we think of a function’s independent variable as representing time (measured, say, in years) we can ask how long it takes for the function’s output to double from some initial value.

Consider $ y = t^{3} $ with an initial value of 8. How many years will pass before its value reaches, say, 16? Well, the function’s output is 8 when t = 2 years, and its output will be 16 when $ t = $ years. Thus, it takes about 0.52 years for the function’s value to double from 8 to 16.

The following exercises about doubling times will pave the way to a surprising development.

Exercises

35. If t is measured in years, how long does it take for….

1. $ y = t^{3} $ to double its value from 1 to 2? From 2 to 4? From 50 to 100?

2. $ y = $ to double its value from 1 to 2? From 2 to 4? From 50 to 100?

3. $ y = t $ to double its value from 1 to 2? From 2 to 4? From 50 to 100?

36. Exponential functions (of the form $ y = ca^{t} $ ) are special with respect to doubling time: the doubling time of any such a function is the same for all initial values. You can see this numerically for yourself: How long does it take for $ y = 2(5^{t}) $ to double its value from 8 to 16? From 16 to 32? From 500 to 1000?

37. Make up your own exponentially growing function of the form $ y = ca^{t} $ , and compute how long it takes for its value to double from 8 to 16, from 250 to 500, and from some number of your choice to twice that number.

38. The numerical evidence you’ve seen in the preceding exercises is compelling, but we can prove definitively that all exponentially growing functions of the form $ y = ca^{t} $ have fixed doubling times, independent of initial value. To prove this, we need logarithms. Your problem: Try to prove it on your own. If, after half an hour or so, you haven’t succeeded, then study the following proof until you understand it.

Claim. If $ f(t) = ca^{t} $ , with a > 1 (to ensure exponential growth), f has a fixed doubling time.

Proof. Let v be an arbitrary positive value.

Solving $ ca^{t}=v $ for t, we see that the function attains the value v when $ t=(v/c)/a $ .

The function’s value doubles when $ ca^{t}=2v $ .

Solving for t again, we see that the function attains the value 2v when $ t = (2v/c) / a $ .

Thus, the time required for f to double its value from v to 2v is

\[ \frac{\ln\left(\frac{2v}{c}\right)}{\ln a}-\frac{\ln\left(\frac{v}{c}\right)}{\ln a}=\frac{\ln\left(\frac{2v}{c}\right)-\ln\left(\frac{v}{c}\right)}{\ln a}=\frac{\ln\left(\frac{\frac{2v}{c}}{\frac{v}{c}}\right)}{\ln a}=\frac{\ln2}{\ln a}.^{*} \]

Note that this expression for doubling time depends only on a; it does not depend upon the initial value v. Thus, the function’s doubling time must be the same for all initial values.

The “Rule of 70”

The “Rule of 70”: If a population grows $ r% $ per year, then it doubles approximately every 70/r years. For example, if an investment earns 7% per year, invested money doubles in about 70/7 = 10 years. (Not bad. Earning 1% interest in a bank, your money would need about 70 years to double.)

To see why this works, consider a population that grows $ r% $ per year. Since $ r% $ means r/100, we know that after t years, the population’s size will be

\[ f(t)=c\left(1+\frac{r}{100}\right)^{t}, \]

where c is the population’s initial size. This is, of course, an exponential function, so according to the proof in exercise 38, its doubling time is

\[ \frac{\ln2}{\ln\left(1+\frac{r}{100}\right)}. \]

Now let’s sacrifice some accuracy for the sake of comprehensibility. Let us approximate this doubling time: We’ll need two facts. First, the numerator is approximately 0.7, as any scientific calculator will confirm. Second, when r is small, we have

\[ \ln\left(1+\frac{r}{100}\right)\approx\frac{r}{100}.^{*} \]

This being so, our function’s doubling time is approximately

\[ \frac{\ln2}{\ln\left(1+\frac{r}{100}\right)}\approx\frac{\ln2}{\frac{r}{100}}\approx\frac{.7}{\frac{r}{100}}=\frac{70}{r}, \]

as claimed.

To approximate that denominator, we had to assume r was “small”. The rule of 70 is well-known in business, since financial growth rates are almost always small enough for that approximation to hold. Take an extreme example: The rule of 70 implies that a 35% growth rate (unbelievably high in business) has a doubling time of 2 years. Even in this extreme case, the rule is still close: The exact doubling time is $ / $ years. For more typical growth rates, the approximation is, of course, better still.

Exercises

39. While walking in the woods with a friend, she mentions that she has heard that the human population of the world, which reached 8 billion in 2022, is growing at 1.1% per year. $ ^{†} $ If this rate of growth continues to hold…

1. In your head (you don’t have a calculator in the woods), use the “rule of 70” to produce a rough estimate as to when world population will reach 16 billion.

2. With the help of a calculator – but without the “rule of 70”, which is just for quick approximations – find a more accurate estimate for when world population will reach 14 billion.

40. For an exponentially growing function of the form $ f(t) = ca^{t} $ , is quadrupling time constant, or does it depend upon the initial value to be quadrupled? What about tripling time?

Half Lives

Clov: Do you believe in the life to come?

Hamm: Mine was always that.

– Samuel Beckett, Endgame

As each exponentially growing function has a fixed doubling time, each exponentially decaying function has a fixed halving time called its “half life”, a name that comes from chemistry. It is best explained with an example.

All living plants and animals contain carbon, which itself comes in several different forms, including the chemically unstable radiocarbon: Over time, radiocarbon “decays” into nitrogen. While a plant or animal lives, its ratio of radiocarbon to ordinary carbon remains constant. But once it dies, and thus ceases to take in new radiocarbon (through breathing or respiration), this ratio begins to decrease, for the radiocarbon in its tissues decays while the ordinary carbon remains chemically stable. Indeed, this ratio decreases by a fixed percentage per year, which means that it decays exponentially, and thus has a fixed half life. The half life of radiocarbon happens to be about 5600 years. $ ^{*} $ Thus, a piece of wood from a tree that died 5600 years ago, or a bone from a man or animal that died 5600 years ago, would have only half as much radiocarbon (proportionately speaking) as their living counterparts.

Radiocarbon dating is a method for determining the ratio of radiocarbon to ordinary carbon in any object made from organic material, and then using this ratio to determine the object’s approximate age. For instance, if we are studying a tool carved by a prehistoric man from a bone or a horn and we find that its radiocarbon-to-ordinary-carbon ratio is only 1/4 of the usual ratio, then, realizing that 1/4 is half of a half, we can conclude that the animal of which it was once part died approximately two “half lives” ago; that is, it must have died about $ 2(5600) = 11,200 $ years ago. On the other hand, if we find that the tool has, say, 40% of the usual radiocarbon-to-ordinary-carbon ratio, then it’s not immediately obvious how many half-lives have passed since the animal died. But… it’s easy to find out. After all, we just need to solve the equation $ (1/2)^{t} = .4 $ . Solving this, we find, as you should verify, that $ t $ . Hence, the animal must have died around $ 1.32(5600) $ years ago.

Radiocarbon dating was developed in the late 1940’s by Willard Libby, who was awarded a Nobel Prize for this work.