Chapter 4

Exponential and Logarithmic Functions

Exponential functions place the variable in the exponent and model real-world phenomena such as population growth, compound interest, and radioactive decay. Their inverses, the logarithmic functions, let us answer questions like when an investment will reach a target or how long a radioactive sample will take to decay to a safe level.

4.1Exponential Functions

Exponential Functions · Graphs of Exponential Functions · Compound Interest

When the variable sits in the exponent, even a small change in the variable can cause a dramatic change in the value of the function. For example, with $ f(x)=2^x $ we have $ f(3)=8 $, $ f(10)=1024 $, and $ f(30)=1{,}073{,}741{,}824 $, while the power function $ g(x)=x^2 $ gives only $ g(30)=900 $.

Exponential Function

The exponential function with base $ a $ is defined for all real numbers $ x $ by $ f(x)=a^x $, where $ a>0 $ and $ a\neq 1 $. We exclude $ a=1 $ because $ 1^x=1 $ is just a constant function. To define $ a^x $ for irrational $ x $, we approximate $ x $ by rational numbers; the Laws of Exponents remain true when the exponents are real numbers.

Exponential Function with Base $ a $

\[ f(x)=a^x, \qquad a>0,\ a\neq 1 \]

EXAMPLE 1 Evaluating Exponential Functions

Let $ f(x)=3^x $. Evaluate $ f(5) $, $ f\!\left(\tfrac{2}{3}\right) $, $ f(\pi) $, and $ f(\sqrt{2}) $.

Solution

$ f(5)=3^5=243 $

$ f\!\left(\tfrac{2}{3}\right)=3^{2/3}\approx 2.0801 $

$ f(\pi)=3^{\pi}\approx 31.5443 $

$ f(\sqrt{2})=3^{\sqrt{2}}\approx 4.7288 $

Graphs of Exponential Functions

Shape and Behavior

The exponential function $ f(x)=a^x $ has domain $ \mathbb{R} $ and range $ (0,\infty) $. The line $ y=0 $ (the x-axis) is a horizontal asymptote, and every graph passes through $ (0,1) $. If $ a>1 $ the function increases rapidly; if \( 0

EXAMPLE 2 Graphing by Plotting Points

Draw the graphs of $ f(x)=3^x $ and $ g(x)=\left(\tfrac{1}{3}\right)^x $.

Solution

Compute several values: for $ f $, $ f(-1)=\tfrac13 $, $ f(0)=1 $, $ f(1)=3 $, $ f(2)=9 $.

Notice that $ g(x)=\left(\tfrac{1}{3}\right)^x=\dfrac{1}{3^x}=3^{-x}=f(-x) $.

So the graph of $ g $ is the graph of $ f $ reflected in the y-axis.

EXAMPLE 3 Identifying Graphs of Exponential Functions

Find the exponential function $ f(x)=a^x $ whose graph passes through (a) $ (2,25) $ and (b) $ \left(3,\tfrac{1}{8}\right) $.

Solution

(a) Since $ f(2)=a^2=25 $, the base is $ a=5 $. So $ f(x)=5^x $.

(b) Since $ f(3)=a^3=\tfrac{1}{8} $, the base is $ a=\tfrac{1}{2} $. So $ f(x)=\left(\tfrac{1}{2}\right)^x $.

EXAMPLE 4 Transformations of Exponential Functions

Use the graph of $ f(x)=2^x $ to sketch (a) $ g(x)=1+2^x $, (b) $ h(x)=-2^x $, (c) $ k(x)=2^{x-1} $. State domain, range, and asymptote.

Solution

(a) Shift upward 1 unit: domain $ \mathbb{R} $, range $ (1,\infty) $, asymptote $ y=1 $.

(b) Reflect in the x-axis: domain $ \mathbb{R} $, range $ (-\infty,0) $, asymptote $ y=0 $.

EXAMPLE 5 Comparing Exponential and Power Functions

Compare the growth of $ f(x)=2^x $ and $ g(x)=x^2 $.

Solution

For small $ x $, $ g(x)=x^2 $ catches up with and overtakes $ f(x)=2^x $ near $ x=2 $.

In a larger window, $ f(x)=2^x $ overtakes $ g(x)=x^2 $ again near $ x=4 $.

For large $ x $, $ 2^x $ is much larger than $ x^2 $: exponential growth eventually dominates any power.

Compound Interest

If a principal $ P $ earns interest at rate $ i $ per period, then after $ k $ periods the amount is $ A=P(1+i)^k $, an exponential function with base $ 1+i $. With annual rate $ r $ compounded $ n $ times per year, the per-period rate is $ r/n $ and there are $ nt $ periods in $ t $ years.

Compound Interest

\[ A(t)=P\left(1+\frac{r}{n}\right)^{nt} \]

$ A(t) $ = amount after $ t $ years, $ P $ = principal, $ r $ = annual rate, $ n $ = compounding periods per year, $ t $ = years.

EXAMPLE 6 Calculating Compound Interest

$1000 is invested at 12% per year. Find the amount after 3 years for several compounding frequencies.

Solution

Use $ A(t)=P\left(1+\tfrac{r}{n}\right)^{nt} $ with $ P=1000 $, $ r=0.12 $, $ t=3 $:

Compounding	$ n $	Amount after 3 yr
Annual	1	$1404.93
Semiannual	2	$1418.52
Quarterly	4	$1425.76
Monthly	12	$1430.77
Daily	365	$1433.24

Annual Percentage Yield (APY)

The annual percentage yield is the simple interest rate that yields the same amount at the end of one year as the given compound interest.

EXAMPLE 7 Calculating the Annual Percentage Yield

Find the APY for an investment earning 6% per year, compounded daily.

Solution

After one year: $ A=P\left(1+\tfrac{0.06}{365}\right)^{365}=P(1.06183) $.

Compare to simple interest $ A=P(1+r) $: so $ 1+r=1.06183 $, giving $ r=0.06183 $.

The annual percentage yield is $ 6.183\% $.

Instructor Note

Instructor Note: Add your section-specific insight here. (Placeholder ready for notes on exponential functions and compound interest.)

4.2The Natural Exponential Function

The Number e · The Natural Exponential Function · Continuously Compounded Interest

The Number $ e $

The number $ e $ is the value that $ \left(1+\tfrac{1}{n}\right)^n $ approaches as $ n $ becomes large. It is irrational, with $ e\approx 2.71828182845904523536 $. The notation $ e $ was chosen by Leonhard Euler, likely for the first letter of "exponential."

The Natural Exponential Function

The natural exponential function uses base $ e $. Since \( 2

Natural Exponential Function

\[ f(x)=e^x \]

EXAMPLE 1 Evaluating the Exponential Function

Evaluate to five decimal places: (a) $ e^3 $, (b) $ 2e^{-0.53} $, (c) $ e^{4.8} $.

Solution

(a) $ e^3\approx 20.08554 $

(b) $ 2e^{-0.53}\approx 1.17721 $

EXAMPLE 2 Graphing the Exponential Functions

Sketch (a) $ f(x)=-e^x $ and (b) $ g(x)=3e^{0.5x} $. State domain, range, asymptote.

Solution

(a) Reflect $ y=e^x $ in the x-axis: domain $ \mathbb{R} $, range $ (-\infty,0) $, asymptote $ y=0 $.

(b) Plot points and connect smoothly: domain $ \mathbb{R} $, range $ (0,\infty) $, asymptote $ y=0 $.

EXAMPLE 3 An Exponential Model for the Spread of a Virus

A disease spreads through a city of 10,000. After $ t $ days the number infected is $ v(t)=\dfrac{10{,}000}{5+1245e^{-0.97t}} $. Find the initial number infected and the counts after 1, 2, and 5 days.

Solution

At $ t=0 $: $ v(0)=\dfrac{10{,}000}{5+1245}=\dfrac{10{,}000}{1250}=8 $ people.

After 1 day: $ v(1)\approx 21 $; after 2 days: $ v(2)\approx 54 $; after 5 days: $ v(5)\approx 678 $.

The graph is a logistic curve: slow rise, rapid growth between days 3 and 8, then leveling off near 2000.

Continuously Compounded Interest

Continuous Compounding

As the number of compounding periods $ n $ increases without bound, the quantity $ \left(1+\tfrac{1}{m}\right)^m $ approaches $ e $, so the compound interest amount approaches $ Pe^{rt} $ — the amount when interest is compounded "at every instant."

Continuously Compounded Interest

\[ A(t)=Pe^{rt} \]

$ A(t) $ = amount after $ t $ years, $ P $ = principal, $ r $ = annual rate, $ t $ = years.

EXAMPLE 4 Calculating Continuously Compounded Interest

Find the amount after 3 years if $1000 is invested at 12% per year, compounded continuously.

Solution

Use $ A(t)=Pe^{rt} $ with $ P=1000 $, $ r=0.12 $, $ t=3 $:

$ A(3)=1000e^{(0.12)3}=1000e^{0.36}\approx \$1433.33 $.

Instructor Note

Instructor Note: Add your section-specific insight here. (Placeholder ready for notes on the number e and continuous compounding.)

4.3Logarithmic Functions

Logarithmic Functions · Graphs · Common Logarithms · Natural Logarithms

Definition of the Logarithmic Function

Every exponential function $ f(x)=a^x $ (with $ a>0 $, $ a\neq 1 $) is one-to-one and so has an inverse, the logarithmic function with base $ a $, written $ \log_a $. In short, $ \log_a x $ is the exponent to which the base $ a $ must be raised to give $ x $.

Logarithmic Function with Base $ a $

\[ \log_a x = y \iff a^y = x \]

EXAMPLE 1 Logarithmic and Exponential Forms

Switch between equivalent logarithmic and exponential forms.

Solution

$ \log_{10} 100{,}000 = 5 \iff 10^5 = 100{,}000 $

$ \log_2 8 = 3 \iff 2^3 = 8 $

$ \log_2 \tfrac{1}{8} = -3 \iff 2^{-3} = \tfrac{1}{8} $

$ \log_5 s = r \iff 5^r = s $

EXAMPLE 2 Evaluating Logarithms

Evaluate each logarithm.

Solution

(a) $ \log_{10} 1000 = 3 $ because $ 10^3 = 1000 $

(b) $ \log_2 32 = 5 $ because $ 2^5 = 32 $

(d) $ \log_{16} 4 = \tfrac{1}{2} $ because $ 16^{1/2} = 4 $

Properties of Logarithms

\[ \log_a 1 = 0 \qquad \log_a a = 1 \qquad \log_a a^x = x \qquad a^{\log_a x} = x \]

EXAMPLE 3 Applying Properties of Logarithms

Illustrate the properties with base 5.

Solution

$ \log_5 1 = 0 $ and $ \log_5 5 = 1 $

$ \log_5 5^8 = 8 $ and $ 5^{\log_5 12} = 12 $

Graphs of Logarithmic Functions

Graph of $ \log_a $

Since $ f(x)=a^x $ has domain $ \mathbb{R} $ and range $ (0,\infty) $, its inverse $ \log_a x $ has domain $ (0,\infty) $ and range $ \mathbb{R} $. The graph is the reflection of $ y=a^x $ in the line $ y=x $. Its x-intercept is 1, and the y-axis is a vertical asymptote.

EXAMPLE 4 Graphing by Plotting Points

Sketch $ f(x)=\log_2 x $.

Solution

Choose x-values that are powers of 2: $ \log_2 \tfrac18=-3 $, $ \log_2 1=0 $, $ \log_2 2=1 $, $ \log_2 8=3 $.

Plot and connect with a smooth curve; the y-axis is a vertical asymptote.

EXAMPLE 5 Reflecting Graphs of Logarithmic Functions

Sketch (a) $ g(x)=-\log_2 x $ and (b) $ h(x)=\log_2(-x) $. State domain, range, asymptote.

Solution

(a) Reflect in the x-axis: domain $ (0,\infty) $, range $ \mathbb{R} $, asymptote $ x=0 $.

(b) Reflect in the y-axis: domain $ (-\infty,0) $, range $ \mathbb{R} $, asymptote $ x=0 $.

EXAMPLE 6 Shifting Graphs of Logarithmic Functions

Sketch (a) $ g(x)=2+\log_5 x $ and (b) $ h(x)=\log_{10}(x-3) $. State domain, range, asymptote.

Solution

(a) Shift up 2 units: domain $ (0,\infty) $, range $ \mathbb{R} $, asymptote $ x=0 $.

(b) Shift right 3 units: domain $ (3,\infty) $, range $ \mathbb{R} $, asymptote $ x=3 $.

Common and Natural Logarithms

Common Logarithm

The logarithm with base 10 is the common logarithm, denoted by omitting the base: $ \log x = \log_{10} x $.

Natural Logarithm

The logarithm with base $ e $ is the natural logarithm, denoted $ \ln $: $ \ln x = \log_e x $. It is the inverse of the natural exponential function, so $ \ln x = y \iff e^y = x $. The properties carry over: $ \ln 1 = 0 $, $ \ln e = 1 $, $ \ln e^x = x $, and $ e^{\ln x} = x $.

Fechner's Law (Human Response to Stimuli)

\[ S = k \log\!\left(\frac{I}{I_0}\right) \]

$ S $ = subjective intensity, $ I $ = physical intensity, $ I_0 $ = threshold intensity, $ k $ = constant for the stimulus.

EXAMPLE 7 Evaluating Common Logarithms

Use a calculator to evaluate $ f(x)=\log x $ at sample values to sketch the graph.

Solution

$ \log 0.01=-2 $, $ \log 0.5\approx-0.301 $, $ \log 1=0 $, $ \log 5\approx0.699 $, $ \log 10=1 $.

Plot these points and connect with a smooth, slowly increasing curve.

EXAMPLE 8 Common Logarithms and Sound

Loudness in decibels is $ B=10\log\!\left(\tfrac{I}{I_0}\right) $. Find the loudness of a sound whose intensity is 100 times $ I_0 $.

Solution

$ B=10\log\!\left(\tfrac{100I_0}{I_0}\right)=10\log 100 $

$ =10\cdot 2 = 20 $. The loudness is 20 dB.

EXAMPLE 9 Evaluating the Natural Logarithm Function

Evaluate each natural logarithm.

Solution

(a) $ \ln e^8 = 8 $

(b) $ \ln\!\left(\tfrac{1}{e^2}\right)=\ln e^{-2}=-2 $

EXAMPLE 10 Finding the Domain of a Logarithmic Function

Find the domain of $ f(x)=\ln(4-x^2) $.

Solution

$ \ln $ is defined when its argument is positive: $ 4-x^2>0 $.

So $ x^2<4 $, i.e. $ |x|<2 $, giving the domain $ (-2,2) $.

EXAMPLE 11 Drawing the Graph of a Logarithmic Function

Graph $ y=x\ln(4-x^2) $ and find its asymptotes and local extrema.

Solution

The domain is $ (-2,2) $, so the lines $ x=-2 $ and $ x=2 $ are vertical asymptotes.

The local maximum value is about 1.13 at $ x\approx1.15 $; by symmetry the local minimum is about $ -1.13 $ at $ x\approx-1.15 $.

Instructor Note

Instructor Note: Add your section-specific insight here. (Placeholder ready for notes on logarithmic functions.)

4.4Laws of Logarithms

Laws of Logarithms · Expanding and Combining Expressions · Change of Base Formula

Laws of Logarithms

Because logarithms are exponents, the Laws of Exponents give rise to the Laws of Logarithms. Let $ a>0 $, $ a\neq 1 $, and let $ A,B>0 $ with $ C $ any real number. The logarithm of a product is a sum, of a quotient is a difference, and of a power is a multiple.

The Three Laws of Logarithms

\[ \log_a(AB)=\log_a A + \log_a B \] \[ \log_a\!\left(\frac{A}{B}\right)=\log_a A - \log_a B \] \[ \log_a(A^C)=C\log_a A \]

EXAMPLE 1 Using the Laws to Evaluate Expressions

Evaluate (a) $ \log_4 2 + \log_4 32 $, (b) $ \log_2 80 - \log_2 5 $, (c) $ -\tfrac{1}{3}\log 8 $.

Solution

(a) $ \log_4(2\cdot 32)=\log_4 64 = 3 $.

(b) $ \log_2\!\left(\tfrac{80}{5}\right)=\log_2 16 = 4 $.

EXAMPLE 2 Expanding Logarithmic Expressions

Expand (a) $ \log_2(6x) $, (b) $ \log_5(x^3 y^6) $, (c) $ \ln\!\left(\dfrac{ab}{\sqrt[3]{c}}\right) $.

Solution

(a) $ \log_2 6 + \log_2 x $

(b) $ 3\log_5 x + 6\log_5 y $

EXAMPLE 3 Combining Logarithmic Expressions

Combine into a single logarithm: (a) $ 3\log x + \tfrac{1}{2}\log(x+1) $, (b) $ 3\ln s + \tfrac{1}{2}\ln t - 4\ln(t^2+1) $.

Solution

(a) $ \log\!\left(x^3(x+1)^{1/2}\right) $

(b) $ \ln\!\left(\dfrac{s^3\sqrt{t}}{(t^2+1)^4}\right) $

Warning: No Law for Sums

There is no rule for the logarithm of a sum or difference. In general $ \log_a(x+y)\neq \log_a x + \log_a y $. Also $ \dfrac{\log 6}{\log 2}\neq \log\!\left(\tfrac{6}{2}\right) $ and $ (\log_2 x)^3 \neq 3\log_2 x $.

EXAMPLE 4 The Law of Forgetting

If a task is learned at level $ P_0 $, then after time $ t $ (months) the performance $ P $ satisfies $ \log P = \log P_0 - c\log(t+1) $. (a) Solve for $ P $. (b) With a score of 90 and $ c=0.2 $, find the expected score after 2 months and after 1 year.

Solution

(a) Combine the right side: $ \log P = \log\dfrac{P_0}{(t+1)^c} $, so $ P=\dfrac{P_0}{(t+1)^c} $.

After 2 months: $ P=\dfrac{90}{(2+1)^{0.2}}\approx 72 $.

After 1 year: $ P=\dfrac{90}{(12+1)^{0.2}}\approx 54 $.

Change of Base Formula

Change of Base

We can express a logarithm in any base in terms of logarithms in another base. This lets us evaluate logarithms of any base using the $ \log $ or $ \ln $ key on a calculator. In particular, $ \log_b a = \dfrac{1}{\log_a b} $.

Change of Base Formula

\[ \log_b x = \frac{\log_a x}{\log_a b} \]

EXAMPLE 5 Evaluating with the Change of Base Formula

Evaluate (a) $ \log_8 5 $ and (b) $ \log_9 20 $ to five decimal places.

Solution

(a) $ \log_8 5 = \dfrac{\log_{10} 5}{\log_{10} 8}\approx 0.77398 $

(b) $ \log_9 20 = \dfrac{\ln 20}{\ln 9}\approx 1.36342 $

EXAMPLE 6 Graphing with the Change of Base Formula

Graph $ f(x)=\log_6 x $ on a calculator.

Solution

Calculators lack a $ \log_6 $ key, so rewrite: $ f(x)=\log_6 x = \dfrac{\ln x}{\ln 6} $.

Enter this form and graph it.

Instructor Note

Instructor Note: Add your section-specific insight here. (Placeholder ready for notes on the Laws of Logarithms.)

4.5Exponential and Logarithmic Equations

Exponential Equations · Logarithmic Equations · Compound Interest

Exponential Equations

An exponential equation has the variable in the exponent. When both sides can be written with the same base, use the one-to-one property: $ a^x = a^y \implies x = y $. Otherwise, take the logarithm of each side to "bring down the exponent."

Guidelines for Solving Exponential Equations

1. Isolate the exponential expression on one side.

2. Take the logarithm of each side; use the Laws of Logarithms to bring down the exponent.

3. Solve for the variable.

EXAMPLE 1 Exponential Equations

Solve (a) $ 5^x = 125 $, (b) $ 5^{2x} = 5^{x+1} $.

Solution

(a) $ 5^x = 5^3 $, so $ x=3 $.

(b) $ 2x = x+1 $, so $ x=1 $.

EXAMPLE 2 Solving an Exponential Equation

Solve $ 3^{x+2} = 7 $: give the exact solution and a six-decimal approximation.

Solution

Take logs: $ (x+2)\log 3 = \log 7 $.

So $ x = \dfrac{\log 7}{\log 3} - 2 $.

Approximation: $ x \approx -0.228756 $.

EXAMPLE 3 Solving an Exponential Equation

Solve $ 8e^{2x} = 20 $.

Solution

Divide by 8: $ e^{2x} = 2.5 $.

Take $ \ln $: $ 2x = \ln 2.5 $.

So $ x = \dfrac{\ln 2.5}{2} \approx 0.458 $.

EXAMPLE 4 Solving Algebraically and Graphically

Solve $ e^{3-2x} = 4 $.

Solution

Take $ \ln $: $ 3-2x = \ln 4 $.

So $ x = \tfrac{1}{2}(3-\ln 4) \approx 0.807 $.

Graphically, $ y=e^{3-2x} $ and $ y=4 $ intersect near $ x\approx 0.81 $.

EXAMPLE 5 An Exponential Equation of Quadratic Type

Solve $ e^{2x} - e^x - 6 = 0 $.

Solution

Treat as a quadratic in $ e^x $: $ (e^x-3)(e^x+2)=0 $.

$ e^x=3 $ gives $ x=\ln 3 $; $ e^x=-2 $ is impossible since $ e^x>0 $.

So $ x=\ln 3 \approx 1.0986 $.

EXAMPLE 6 An Equation Involving Exponential Functions

Solve $ 3xe^x + x^2 e^x = 0 $.

Solution

Factor: $ x(3+x)e^x = 0 $.

Divide by $ e^x $ (never zero): $ x(3+x)=0 $.

So $ x=0 $ or $ x=-3 $.

Logarithmic Equations

A logarithmic equation contains a logarithm of the variable. Use the one-to-one property $ \log_a x = \log_a y \implies x = y $, or rewrite the equation in exponential form. Always check solutions, since logarithms of nonpositive numbers are undefined.

Guidelines for Solving Logarithmic Equations

1. Isolate the logarithmic term, combining logs if necessary.

2. Rewrite in exponential form (or raise the base to each side).

3. Solve for the variable, then check.

EXAMPLE 7 Solving a Logarithmic Equation

Solve $ \log_5(x^2-1) = \log_5(x-2) + \log_5(x+3) $.

Solution

Combine the right side: $ \log_5(x^2-1) = \log_5(x^2+x-6) $.

By one-to-one: $ x^2-1 = x^2+x-6 $.

So $ x=7 $ (which checks).

EXAMPLE 8 Solving Logarithmic Equations

Solve (a) $ \ln x = 8 $ and (b) $ \log_2(25-x) = 3 $.

Solution

(a) Exponential form: $ x = e^8 \approx 2981 $.

(b) $ 25-x = 2^3 = 8 $, so $ x = 17 $.

EXAMPLE 9 Solving a Logarithmic Equation

Solve $ 4 + 3\log(2x) = 16 $.

Solution

Isolate: $ 3\log(2x)=12 $, so $ \log(2x)=4 $.

Exponential form: $ 2x = 10^4 $.

So $ x = 5000 $.

EXAMPLE 10 Solving Algebraically and Graphically

Solve $ \log(x+2) + \log(x-1) = 1 $.

Solution

Combine: $ \log[(x+2)(x-1)] = 1 $, so $ (x+2)(x-1)=10 $.

Expand and solve: $ x^2+x-12=0 $, giving $ (x+4)(x-3)=0 $.

$ x=-4 $ is rejected (negative argument); $ x=3 $ is the only solution.

EXAMPLE 11 Solving a Logarithmic Equation Graphically

Solve $ x^2 = 2\ln(x+2) $.

Solution

It cannot be isolated algebraically, so graph $ y = x^2 - 2\ln(x+2) $.

The x-intercepts give two solutions: $ x\approx -0.71 $ and $ x\approx 1.60 $.

EXAMPLE 12 Transparency of a Lake (Beer–Lambert Law)

The Beer–Lambert Law gives $ -\tfrac{1}{k}\ln\!\left(\tfrac{I}{I_0}\right)=x $. (a) Solve for $ I $. (b) With $ k=0.025 $ and $ I_0=14 $ lm, find the intensity at depth 20 ft.

Solution

(a) Multiply by $ -k $: $ \ln\!\left(\tfrac{I}{I_0}\right) = -kx $. Exponential form: $ I = I_0 e^{-kx} $.

(b) $ I = 14e^{-(0.025)(20)} \approx 8.49 $, so about 8.5 lm.

Compound Interest

Interest Formulas (Summary)

\[ A = P(1+r) \quad\text{(simple, one year)} \] \[ A(t) = P\left(1+\frac{r}{n}\right)^{nt} \quad\text{(}n\text{ times per year)} \] \[ A(t) = Pe^{rt} \quad\text{(continuous)} \]

EXAMPLE 13 Finding the Term for an Investment to Double

$5000 is invested at 5% per year. Find the time to double if compounded (a) semiannually, (b) continuously.

Solution

(a) $ (1.025)^{2t}=2 $, so $ t=\dfrac{\log 2}{2\log 1.025}\approx 14.04 $ years.

(b) $ e^{0.05t}=2 $, so $ t=\dfrac{\ln 2}{0.05}\approx 13.86 $ years.

EXAMPLE 14 Time Required to Grow an Investment

$1000 is invested at 4% per year, compounded continuously. Find the time for it to grow to $4000.

Solution

$ 1000e^{0.04t}=4000 $, so $ e^{0.04t}=4 $.

$ 0.04t=\ln 4 $, giving $ t=\dfrac{\ln 4}{0.04}\approx 34.66 $ years (about 34 years 8 months).

Instructor Note

Instructor Note: Add your section-specific insight here. (Placeholder ready for notes on solving exponential and logarithmic equations.)

4.6Modeling with Exponential Functions

Exponential Growth (Doubling Time) · Exponential Growth (Relative Growth Rate) · Radioactive Decay · Newton's Law of Cooling

Exponential Growth (Doubling Time)

If a population has initial size $ n_0 $ and doubling time $ a $, then the size at time $ t $ is modeled with base 2, where $ a $ and $ t $ are measured in the same units.

Exponential Growth (Doubling Time)

\[ n(t)=n_0\, 2^{t/a} \]

EXAMPLE 1 Bacteria Population

A bacteria population doubles every 3 hours, starting at 1000. (a) Model it. (b) Find the count after 15 hours. (c) When does it reach 100,000?

Solution

(a) $ n(t)=1000\cdot 2^{t/3} $.

(b) $ n(15)=1000\cdot 2^{5}=32{,}000 $.

EXAMPLE 2 Rabbit Population

Rabbits introduced 8 months ago number 4100 and double every 3 months. (a) Find the initial size. (b) Estimate the population after 1 year.

Solution

(a) $ 4100=n_0\,2^{8/3} $, so $ n_0=\dfrac{4100}{2^{8/3}}\approx 645 $.

(b) $ n(12)=645\cdot 2^{12/3}=645\cdot 2^4 = 10{,}320 $, about 10,000 rabbits.

Exponential Growth (Relative Growth Rate)

Using base $ e $, a population is modeled by the relative growth rate $ r $ — the instantaneous rate expressed as a proportion of the current population. This is the same principle as continuously compounded interest.

Exponential Growth (Relative Growth Rate)

\[ n(t)=n_0\, e^{rt} \]

$ n_0 $ = initial size, $ r $ = relative growth rate, $ t $ = time.

EXAMPLE 3 Predicting the Size of a Population

A culture starts at 500 bacteria with relative growth rate 40% per hour. (a) Model it. (b) Find the count after 10 hours. (c) When does it reach 80,000?

Solution

(a) $ n(t)=500e^{0.4t} $.

(b) $ n(10)=500e^{4}\approx 27{,}300 $.

EXAMPLE 4 Comparing Different Rates of Population Growth

In 2000 world population was 6.1 billion. Estimate it in 2050 using growth rates of (a) 1.4% and (b) 1.0% per year.

Solution

(a) $ n(50)=6.1e^{0.014(50)}=6.1e^{0.7}\approx 12.3 $ billion.

(b) $ n(50)=6.1e^{0.010(50)}=6.1e^{0.5}\approx 10.1 $ billion.

A small change in rate makes a large difference over time.

EXAMPLE 5 Expressing a Model in Terms of e

A culture starts with 10,000 bacteria and doubles every 40 minutes. Express the model with base 2 and with base $ e $ (time in hours).

Solution

Doubling time $ a=40 $ min $ =\tfrac{2}{3} $ h, so $ \tfrac{1}{a}=1.5 $: $ n(t)=10{,}000\cdot 2^{1.5t} $.

Find $ r $ from $ 20{,}000=10{,}000e^{r(2/3)} $: $ r=\dfrac{3\ln 2}{2}\approx 1.0397 $.

So $ n(t)=10{,}000\,e^{1.0397t} $. Both models give identical graphs.

Radioactive Decay

Radioactive Decay Model

Radioactive substances decay at a rate proportional to their mass. The half-life $ h $ is the time for a sample to decay to half its mass. With base $ e $, the relative decay rate is found from the half-life as $ r=\dfrac{\ln 2}{h} $.

Radioactive Decay Model

\[ m(t)=m_0\, 2^{-t/h} \qquad\text{or}\qquad m(t)=m_0\, e^{-rt},\quad r=\frac{\ln 2}{h} \]

EXAMPLE 6 Radioactive Decay

Polonium-210 has half-life 140 days; a sample is 300 mg. (a)–(b) Model the mass. (c) Find the mass after one year. (d) When does it decay to 200 mg?

Solution

(a) $ m(t)=300\cdot 2^{-t/140} $.

(b) $ r=\dfrac{\ln 2}{140}\approx 0.00495 $, so $ m(t)=300e^{-0.00495t} $.

(d) Solve $ 300e^{-0.00495t}=200 $: $ t=-\dfrac{\ln(2/3)}{0.00495}\approx 82 $ days.

Newton's Law of Cooling

The rate at which an object cools is proportional to the temperature difference between the object and its surroundings (when that difference is not too large). The surrounding temperature $ T_s $ is a horizontal asymptote of the temperature function.

Newton's Law of Cooling

\[ T(t)=T_s + D_0\, e^{-kt} \]

$ T_s $ = surrounding temperature, $ D_0 $ = initial temperature difference, $ k>0 $ = constant for the object.

EXAMPLE 7 Newton's Law of Cooling

Coffee at 200°F is placed in a 70°F room; after 10 min it is 150°F. (a) Model the temperature. (b) Find it after 15 min. (c) When does it reach 100°F?

Solution

(a) $ D_0=200-70=130 $. Use $ T(10)=150 $: $ 130e^{-10k}=80 $, so $ k=-\tfrac{1}{10}\ln\tfrac{8}{13}\approx 0.04855 $. Thus $ T(t)=70+130e^{-0.04855t} $.

(b) $ T(15)=70+130e^{-0.04855(15)}\approx 133 $°F.

Instructor Note

Instructor Note: Add your section-specific insight here. (Placeholder ready for notes on exponential models.)

4.7Logarithmic Scales

The pH Scale · The Richter Scale · The Decibel Scale

Logarithmic Scales

When a quantity varies over a very large range, taking its logarithm produces more manageable numbers. On a logarithmic scale, numbers are represented by their logarithms — making it easy to compare values that span many orders of magnitude.

The pH Scale

The pH measures acidity from the hydrogen ion concentration $ [\mathrm{H}^+] $ (in moles per liter). Solutions with pH 7 are neutral; pH below 7 is acidic and above 7 is basic. When pH increases by one unit, $ [\mathrm{H}^+] $ decreases by a factor of 10.

The pH Scale

\[ \mathrm{pH} = -\log[\mathrm{H}^+] \]

EXAMPLE 1 pH and Hydrogen Ion Concentration

(a) Blood has $ [\mathrm{H}^+]=3.16\times 10^{-8} $ M. Find its pH and classify it. (b) Rainfall with pH 2.4 — find $ [\mathrm{H}^+] $.

Solution

(a) $ \mathrm{pH}=-\log(3.16\times 10^{-8})\approx 7.5 $; since it exceeds 7, the blood is basic.

(b) $ [\mathrm{H}^+]=10^{-\mathrm{pH}}=10^{-2.4}\approx 4.0\times 10^{-3} $ M.

The Richter Scale

The magnitude $ M $ of an earthquake compares its intensity $ I $ to that of a standard earthquake $ S $. A standard earthquake has magnitude $ M=\log\tfrac{S}{S}=\log 1=0 $. An increase of one unit on the scale corresponds to a tenfold increase in intensity.

The Richter Scale

\[ M = \log\frac{I}{S} \]

EXAMPLE 2 Magnitude and Intensity

(a) Find the magnitude of a quake with intensity 3.75 (with $ S=10^{-4} $). (b) A quake measures 5.1 — find its intensity.

Solution

(a) $ M=\log\dfrac{3.75}{10^{-4}}=\log 37500\approx 4.6 $.

(b) From $ 10^M=\dfrac{I}{S} $: $ I=10^{5.1}\cdot 10^{-4}=10^{1.1}\approx 12.6 $.

EXAMPLE 3 Magnitude of Earthquakes

The 1906 San Francisco quake had magnitude 8.3. A quake four times as intense occurred the same year. Find its magnitude.

Solution

For intensity $ 4I $: $ M=\log\dfrac{4I}{S}=\log 4 + \log\dfrac{I}{S} $.

$ =\log 4 + 8.3 \approx 8.9 $.

EXAMPLE 4 Intensity of Earthquakes

The 1989 Loma Prieta quake had magnitude 7.1. How many times more intense was the 1906 quake (magnitude 8.3)?

Solution

$ \log\dfrac{I_1}{I_2}=\log\dfrac{I_1}{S}-\log\dfrac{I_2}{S}=8.3-7.1=1.2 $.

So $ \dfrac{I_1}{I_2}=10^{1.2}\approx 16 $: the 1906 quake was about 16 times as intense.

The Decibel Scale

The decibel level $ B $ compares a sound's intensity $ I $ to the reference $ I_0=10^{-12} $ W/m² (the threshold of hearing). Perceived loudness varies with the logarithm of intensity (the Weber–Fechner Law). The threshold of pain is about 120 dB.

The Decibel Scale

\[ B = 10\log\frac{I}{I_0} \]

EXAMPLE 5 Decibel Level and Intensity

(a) Find the decibel level of a jet engine with intensity $ 10^2 $ W/m². (b) A motorcycle measures 90 dB — find its intensity.

Solution

(a) $ B=10\log\dfrac{10^2}{10^{-12}}=10\log 10^{14}=140 $ dB.

(b) $ \dfrac{90}{10}=\log I + 12 $, so $ \log I = -3 $ and $ I = 10^{-3} $ W/m².

Instructor Note

Instructor Note: Add your section-specific insight here. (Placeholder ready for notes on logarithmic scales.)

FMFitting Exponential and Power Curves to Data

Exponential Models · Power Models · Linearizing Data · Logistic Models

Modeling with Exponential Functions

When a scatter plot of data increases rapidly, an exponential model $ f(x)=Ce^{kx} $ (with constants $ C $ and $ k $) may fit. A graphing calculator's exponential regression gives a model of the form $ y=Cb^t $, which converts to base $ e $ using $ b^t = e^{t\ln b} $.

Exponential Model

\[ f(x)=Ce^{kx} \]

EXAMPLE 1 An Exponential Model for World Population

Given 20th-century world population data, find an exponential model and predict population in 2020.

Solution

Regression gives $ P(t)=(0.0082543)(1.0137186)^t $.

Convert to base $ e $ using $ \ln 1.0137186\approx 0.013625 $: $ P(t)=0.0082543e^{0.013625t} $.

Predicted 2020: $ P(2020)\approx 7{,}405{,}400{,}000 $.

Modeling with Power Functions

If a scatter plot resembles a power curve, use a power model $ f(x)=ax^n $, where $ a>0 $ and $ n $ is any real number. Power regression on a calculator finds the best-fit values of $ a $ and $ n $.

Power Model

\[ f(x)=ax^n \]

EXAMPLE 2 A Power Model for Planetary Periods

Given each planet's mean distance $ d $ (AU) and period $ T $ (years), find a power model, then find the period of an asteroid at $ d=5 $ AU.

Solution

Power regression gives $ T=1.000396\,d^{1.49966} $, which rounds to Kepler's relation $ T=d^{1.5} $.

At $ d=5 $: $ T=1.00039\cdot 5^{1.49966}\approx 11.22 $ years.

Linearizing Data

To decide between an exponential and a power model, linearize the data. For exponential data $ y=Ce^{kx} $, the points $ (x,\ln y) $ lie on a line (a semi-log plot). For power data $ y=ax^n $, the points $ (\ln x,\ln y) $ lie on a line (a log-log plot). Whichever plot is more nearly straight indicates the appropriate model.

Linearized Relationships

\[ y=Ce^{kx} \ \Rightarrow\ \ln y = kx + \ln C \quad\text{(semi-log: line)} \] \[ y=ax^n \ \Rightarrow\ \ln y = n\ln x + \ln a \quad\text{(log-log: line)} \]

EXAMPLE 3 An Exponential or Power Model?

For a set of data points, decide whether an exponential or power function is appropriate, then find it.

Solution

The semi-log plot $ (x,\ln y) $ is not linear, so exponential is inappropriate.

The log-log plot $ (\ln x,\ln y) $ is nearly linear, so a power model fits.

Power regression gives $ y=1.85x^{1.82} $.

Modeling with Logistic Functions

A logistic growth model has the form below, with positive constants $ a $, $ b $, $ c $. Logistic functions model populations whose growth is constrained by limited resources. As $ t\to\infty $, $ e^{-bt}\to 0 $, so the population approaches the carrying capacity $ c $.

Logistic Growth Model

\[ f(t)=\frac{c}{1+ae^{-bt}} \]

EXAMPLE 4 Stocking a Pond with Catfish

A pond is stocked with 1000 catfish and sampled every 15 weeks. Find a model and describe the growth.

Solution

Logistic regression gives $ P(t)=\dfrac{7925}{1+7.7e^{-0.052t}} $.

The population rises rapidly until about week 80, then levels off near 7900 — the carrying capacity.

Instructor Note

Instructor Note: Add your section-specific insight here. (Placeholder ready for notes on curve fitting and regression models.)

Exponential and Logarithmic Functions

4.1Exponential Functions

Exponential Function

Graphs of Exponential Functions

Shape and Behavior

Compound Interest

Compound Interest

Annual Percentage Yield (APY)

4.2The Natural Exponential Function

The Number \( e \)

The Natural Exponential Function

Continuously Compounded Interest

Continuous Compounding

4.3Logarithmic Functions

Definition of the Logarithmic Function

Graphs of Logarithmic Functions

Graph of \( \log_a \)

Common and Natural Logarithms

Common Logarithm

Natural Logarithm

4.4Laws of Logarithms

Laws of Logarithms

Warning: No Law for Sums

Change of Base Formula

Change of Base

4.5Exponential and Logarithmic Equations

Exponential Equations

Logarithmic Equations

Logarithmic Equations

Compound Interest

4.6Modeling with Exponential Functions

Exponential Growth (Doubling Time)

Exponential Growth (Relative Growth Rate)

Radioactive Decay

Radioactive Decay Model

Newton's Law of Cooling

Newton's Law of Cooling

4.7Logarithmic Scales

Logarithmic Scales

The pH Scale

The pH Scale

The Richter Scale

The Richter Scale

The Decibel Scale

The Decibel Scale

FMFitting Exponential and Power Curves to Data

Modeling with Exponential Functions

Modeling with Power Functions

Linearizing Data

Modeling with Logistic Functions