Chapter 6: Exponential & Logarithmic Functions
Exponential and logarithmic functions are among the most important tools in mathematics. They model phenomena that grow or shrink at rates proportional to their current size: populations doubling every generation, radioactive substances losing half their mass at regular intervals, investments compounding over time. In this chapter we build a thorough understanding of these functions, learn how to manipulate and solve equations involving them, and apply them to real-world problems in finance, biology, chemistry, and physics.
6.1 Exponential Functions
Up to this point in Algebra 2, the functions we have studied—linear, quadratic, polynomial—have the variable $x$ appearing in the base with a constant exponent. Exponential functions reverse this relationship: the variable appears in the exponent.
Definition: Exponential Function
An exponential function is any function of the form
$$f(x) = a \cdot b^x$$
where $a \neq 0$, $b > 0$, and $b \neq 1$. Here $a$ is the initial value (the $y$-intercept when the function is written in this standard form) and $b$ is the base.
The restriction $b > 0$ ensures the function is defined for all real numbers $x$. We also exclude $b = 1$ because $1^x = 1$ for all $x$, which produces the constant function $f(x) = a$ rather than an exponential.
Growth vs. Decay
The behavior of an exponential function depends entirely on whether the base $b$ is greater than or less than 1:
- Exponential Growth ($b > 1$): As $x$ increases, $b^x$ increases without bound. The function rises steeply to the right. Common examples include population growth and compound interest.
- Exponential Decay ($0 < b < 1$): As $x$ increases, $b^x$ approaches 0. The function decreases toward the $x$-axis but never reaches it. Common examples include radioactive decay and depreciation.
Key Properties of $f(x) = a \cdot b^x$ (with $a > 0$)
Domain: $(-\infty, \infty)$ — all real numbers.
Range: $(0, \infty)$ — always positive.
$y$-intercept: $(0, a)$ because $f(0) = a \cdot b^0 = a$.
Horizontal asymptote: $y = 0$ (the $x$-axis).
Monotonicity: Strictly increasing if $b > 1$; strictly decreasing if $0 < b < 1$.
Graphing and Transformations
We can apply the standard toolkit of transformations to exponential functions. Given a parent function $f(x) = b^x$, the general transformed form is:
$$g(x) = a \cdot b^{x - h} + k$$
Each parameter controls a specific transformation:
- $a$: Vertical stretch (if $|a| > 1$) or compression (if $0 < |a| < 1$). If $a < 0$, the graph reflects over the $x$-axis.
- $h$: Horizontal shift. The graph moves $h$ units to the right.
- $k$: Vertical shift. The horizontal asymptote moves from $y = 0$ to $y = k$.
Example 1: Identifying Growth or Decay
Classify each function as exponential growth or exponential decay:
(a) $f(x) = 3 \cdot 2^x$ — Since $b = 2 > 1$, this is exponential growth.
(b) $g(x) = 5 \cdot (0.4)^x$ — Since $b = 0.4$ and $0 < 0.4 < 1$, this is exponential decay.
(c) $h(x) = 100 \cdot \left(\frac{3}{2}\right)^x$ — Since $b = \frac{3}{2} > 1$, this is exponential growth.
The Natural Base $e$
Among all possible bases for exponential functions, one stands out as the most mathematically natural. The number $e$, known as Euler's number, is defined as
$$e = \lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^n \approx 2.71828$$
The function $f(x) = e^x$ has a remarkable property: at every point, its rate of change equals its current value. This makes $e^x$ the natural choice for modeling continuous growth and decay processes. In calculus, you will learn that the derivative of $e^x$ is simply $e^x$ itself—no other base has this property.
Why $e$ Matters: While any base can model exponential growth, $e$ simplifies the mathematics of continuous processes. Whenever a quantity changes at a rate proportional to itself (population growth, radioactive decay, continuously compounded interest), the natural base $e$ is the right tool.
6.2 Logarithmic Functions
Because exponential functions are one-to-one (they pass the horizontal line test), each exponential function has an inverse. That inverse is the logarithmic function.
Definition: Logarithm
For $b > 0$ and $b \neq 1$, the logarithm base $b$ of a positive number $x$ is defined as:
$$\log_b x = y \quad \Longleftrightarrow \quad b^y = x$$
In words: $\log_b x$ is the exponent to which we must raise $b$ to obtain $x$.
This definition establishes a direct correspondence between exponential and logarithmic forms. Being able to convert fluently between these two forms is essential for solving equations.
Example 2: Converting Between Forms
Rewrite each equation in the other form:
(a) $2^5 = 32 \;\Longleftrightarrow\; \log_2 32 = 5$
(b) $\log_3 81 = 4 \;\Longleftrightarrow\; 3^4 = 81$
(c) $10^{-2} = 0.01 \;\Longleftrightarrow\; \log_{10} 0.01 = -2$
(d) $e^0 = 1 \;\Longleftrightarrow\; \ln 1 = 0$
Common Log and Natural Log
Two logarithmic bases appear so frequently that they have special notation:
- Common logarithm: $\log x = \log_{10} x$. Used in pH calculations, decibel scales, and the Richter scale.
- Natural logarithm: $\ln x = \log_e x$. Used throughout calculus, physics, and continuous growth models.
Your calculator has dedicated buttons for both: LOG for $\log_{10}$ and LN for $\ln$.
Key Properties of $f(x) = \log_b x$ (with $b > 1$)
Domain: $(0, \infty)$ — only positive inputs.
Range: $(-\infty, \infty)$ — all real numbers.
$x$-intercept: $(1, 0)$ because $\log_b 1 = 0$.
Vertical asymptote: $x = 0$ (the $y$-axis).
Key point: $(b, 1)$ because $\log_b b = 1$.
Behavior: Increasing, but grows very slowly compared to polynomial or exponential functions.
Since $y = \log_b x$ is the inverse of $y = b^x$, their graphs are reflections of each other across the line $y = x$. This geometric relationship provides immediate insight: any horizontal asymptote of the exponential becomes a vertical asymptote of the logarithm, and vice versa.
Inverse Relationship: Always remember the cancellation identities that follow from the inverse relationship:
$$b^{\log_b x} = x \quad \text{for } x > 0 \qquad \text{and} \qquad \log_b(b^x) = x \quad \text{for all } x$$
6.3 Properties of Logarithms
Because logarithms are exponents, the familiar exponent rules translate directly into logarithm rules. These properties allow us to expand, condense, and simplify logarithmic expressions—skills that are essential for solving equations in the next two sections.
Properties of Logarithms
Let $b > 0$, $b \neq 1$, and let $M, N > 0$. Then:
Product Rule: $\log_b(MN) = \log_b M + \log_b N$
Quotient Rule: $\log_b\!\left(\dfrac{M}{N}\right) = \log_b M - \log_b N$
Power Rule: $\log_b(M^p) = p \cdot \log_b M$ for any real number $p$
These properties follow directly from the corresponding exponent laws. For instance, the product rule mirrors the fact that $b^m \cdot b^n = b^{m+n}$. If $\log_b M = m$ and $\log_b N = n$, then $MN = b^m \cdot b^n = b^{m+n}$, so $\log_b(MN) = m + n = \log_b M + \log_b N$.
Example 3: Expanding Logarithmic Expressions
Expand completely: $\log_2\!\left(\dfrac{x^3 \sqrt{y}}{z^2}\right)$
Solution: Apply the quotient rule, then the product rule, then the power rule:
$$\log_2\!\left(\frac{x^3 \sqrt{y}}{z^2}\right) = \log_2(x^3 \sqrt{y}) - \log_2(z^2)$$
$$= \log_2(x^3) + \log_2(\sqrt{y}) - \log_2(z^2)$$
$$= 3\log_2 x + \frac{1}{2}\log_2 y - 2\log_2 z$$
Example 4: Condensing Logarithmic Expressions
Write as a single logarithm: $2\ln x + 3\ln y - \ln z$
Solution: Apply the power rule in reverse, then the product and quotient rules:
$$2\ln x + 3\ln y - \ln z = \ln x^2 + \ln y^3 - \ln z = \ln\!\left(\frac{x^2 y^3}{z}\right)$$
Change of Base Formula
Since most calculators only compute $\log$ (base 10) and $\ln$ (base $e$), we need a way to evaluate logarithms with any base. The change of base formula provides this.
Change of Base Formula
For any positive numbers $a$, $b$, and $x$ with $a \neq 1$ and $b \neq 1$:
$$\log_b x = \frac{\log_a x}{\log_a b}$$
In practice, using $a = 10$ or $a = e$:
$$\log_b x = \frac{\log x}{\log b} = \frac{\ln x}{\ln b}$$
Example 5: Using the Change of Base Formula
Evaluate $\log_5 40$ to four decimal places.
Solution:
$$\log_5 40 = \frac{\ln 40}{\ln 5} = \frac{3.6889\ldots}{1.6094\ldots} \approx 2.2920$$
6.4 Solving Exponential Equations
An exponential equation is one in which the variable appears in an exponent. There are two main strategies for solving these equations, depending on whether the two sides can be written with the same base.
Strategy 1: Same Base
If both sides of an equation can be expressed as powers of the same base, then we can use the one-to-one property:
One-to-One Property of Exponentials
If $b > 0$ and $b \neq 1$, then:
$$b^m = b^n \quad \Longleftrightarrow \quad m = n$$
Example 6: Same Base
Solve: $8^{x+1} = 4^{2x-3}$
Solution: Express both sides as powers of 2:
$$\left(2^3\right)^{x+1} = \left(2^2\right)^{2x-3}$$
$$2^{3(x+1)} = 2^{2(2x-3)}$$
$$2^{3x+3} = 2^{4x-6}$$
By the one-to-one property: $3x + 3 = 4x - 6$, so $x = 9$.
Strategy 2: Using Logarithms
When the two sides cannot be rewritten with a common base, we take the logarithm of both sides. You can use either $\log$ or $\ln$; both give the same answer.
Example 7: Solving with Logarithms
Solve: $5^{2x} = 17$
Solution: Take the natural logarithm of both sides:
$$\ln(5^{2x}) = \ln 17$$
$$2x \cdot \ln 5 = \ln 17$$
$$x = \frac{\ln 17}{2 \ln 5} = \frac{2.8332\ldots}{2(1.6094\ldots)} \approx 0.8804$$
Example 8: Equation with $e$
Solve: $3e^{4x-1} = 60$
Solution: Isolate the exponential first, then take $\ln$:
$$e^{4x-1} = 20$$
$$4x - 1 = \ln 20$$
$$x = \frac{1 + \ln 20}{4} = \frac{1 + 2.9957\ldots}{4} \approx 0.9989$$
Common Mistake: Students sometimes try to "distribute" a logarithm: $\log(a + b) \neq \log a + \log b$. The product rule says $\log(ab) = \log a + \log b$, but there is no rule for the logarithm of a sum.
6.5 Solving Logarithmic Equations
A logarithmic equation contains a logarithm of an expression involving the variable. The key strategy is to convert from logarithmic form to exponential form, or to use properties of logarithms to combine multiple logarithms into one.
Strategy 1: Convert to Exponential Form
Example 9: Direct Conversion
Solve: $\log_3(2x + 1) = 4$
Solution: Convert to exponential form:
$$3^4 = 2x + 1$$
$$81 = 2x + 1$$
$$x = 40$$
Check domain: $2(40) + 1 = 81 > 0$. Valid.
Strategy 2: Use Properties to Combine
When an equation has multiple logarithmic terms, use the product, quotient, or power rules to combine them into a single logarithm, then convert to exponential form.
Example 10: Combining Logarithms
Solve: $\log(x + 3) + \log(x - 2) = \log 14$
Solution: Use the product rule on the left side:
$$\log\big[(x + 3)(x - 2)\big] = \log 14$$
Since $\log$ is one-to-one: $(x + 3)(x - 2) = 14$
$$x^2 + x - 6 = 14$$
$$x^2 + x - 20 = 0$$
$$(x + 5)(x - 4) = 0$$
$$x = -5 \quad \text{or} \quad x = 4$$
Check domain: For $x = -5$: $\log(-5 + 3) = \log(-2)$ is undefined. Reject.
For $x = 4$: $\log(7)$ and $\log(2)$ are both defined. $\log(7 \cdot 2) = \log 14$. Valid.
$$\boxed{x = 4}$$
Always Check Domain: Every solution to a logarithmic equation must be checked against the original equation. The argument of every logarithm must be positive. Extraneous solutions that violate this condition must be rejected.
Example 11: Natural Log Equation
Solve: $\ln(x^2) = \ln(5x + 6)$
Solution: Since $\ln$ is one-to-one:
$$x^2 = 5x + 6$$
$$x^2 - 5x - 6 = 0$$
$$(x - 6)(x + 1) = 0$$
$$x = 6 \quad \text{or} \quad x = -1$$
Check: For $x = 6$: $\ln(36) = \ln(36)$. Valid.
For $x = -1$: $\ln(1) = \ln(1)$. Valid (since $(-1)^2 = 1 > 0$ and $5(-1) + 6 = 1 > 0$).
$$\boxed{x = 6 \text{ or } x = -1}$$
6.6 Applications
Exponential and logarithmic functions are not abstract curiosities—they model some of the most important quantitative relationships in the world. In this section we explore four major application areas.
Compound Interest
Compound Interest Formula
If a principal amount $P$ is invested at an annual interest rate $r$ (as a decimal), compounded $n$ times per year, then after $t$ years the balance is:
$$A = P\!\left(1 + \frac{r}{n}\right)^{nt}$$
The more frequently interest is compounded, the more the investment grows. As $n \to \infty$, we arrive at continuous compounding:
Continuous Compounding Formula
$$A = Pe^{rt}$$
This represents the theoretical maximum growth for a given rate, where interest is compounded at every instant.
Example 12: Comparing Compounding Frequencies
You invest $\$5{,}000$ at $6\%$ annual interest for 10 years. Compare the final amounts under different compounding frequencies.
Annually ($n = 1$): $A = 5000(1.06)^{10} \approx \$8{,}954.24$
Quarterly ($n = 4$): $A = 5000\left(1 + \frac{0.06}{4}\right)^{40} \approx \$9{,}070.09$
Monthly ($n = 12$): $A = 5000\left(1 + \frac{0.06}{12}\right)^{120} \approx \$9{,}096.98$
Continuously: $A = 5000 \cdot e^{0.6} \approx \$9{,}110.59$
Continuous compounding yields $\$156.35$ more than annual compounding over 10 years.
Example 13: Doubling Time
How long does it take for an investment to double at $8\%$ annual interest, compounded continuously?
Solution: We need $2P = Pe^{0.08t}$, so $2 = e^{0.08t}$.
$$\ln 2 = 0.08t$$
$$t = \frac{\ln 2}{0.08} = \frac{0.6931}{0.08} \approx 8.66 \text{ years}$$
Rule of 72: A quick approximation for doubling time: divide 72 by the interest rate percentage. At $8\%$: $72 \div 8 = 9$ years, which is close to the exact answer of 8.66 years.
Population Growth
Many biological populations grow exponentially when resources are plentiful. The standard model is:
$$P(t) = P_0 \cdot e^{kt}$$
where $P_0$ is the initial population and $k$ is the growth rate constant. If $k > 0$, the population is growing; if $k < 0$, it is declining.
Example 14: Bacterial Growth
A colony of bacteria contains 800 cells at time $t = 0$ and 2400 cells after 3 hours. Assuming exponential growth, find (a) the growth rate $k$ and (b) the population after 8 hours.
Solution:
(a) $2400 = 800 \cdot e^{3k}$, so $3 = e^{3k}$ and $k = \dfrac{\ln 3}{3} \approx 0.3662$.
(b) $P(8) = 800 \cdot e^{0.3662 \cdot 8} = 800 \cdot e^{2.9296} \approx 800 \cdot 18.72 \approx 14{,}976$ cells.
Radioactive Decay and Half-Life
Radioactive substances decay at a rate proportional to the amount present, following the model:
$$A(t) = A_0 \cdot e^{-kt}$$
The half-life $t_{1/2}$ is the time required for half the substance to decay. Setting $A(t_{1/2}) = \frac{1}{2}A_0$:
$$\frac{1}{2} = e^{-kt_{1/2}} \quad \Longrightarrow \quad t_{1/2} = \frac{\ln 2}{k}$$
Example 15: Carbon-14 Dating
Carbon-14 has a half-life of approximately 5730 years. A fossil contains $35\%$ of its original carbon-14. How old is it?
Solution: First find $k$: $k = \dfrac{\ln 2}{5730} \approx 0.0001210$
Then solve $0.35 = e^{-0.0001210\, t}$:
$$\ln(0.35) = -0.0001210\, t$$
$$t = \frac{-\ln(0.35)}{0.0001210} = \frac{1.0498}{0.0001210} \approx 8{,}678 \text{ years}$$
The pH Scale
The pH of a solution is a logarithmic measure of its hydrogen ion concentration $[\text{H}^+]$, measured in moles per liter:
$$\text{pH} = -\log[\text{H}^+]$$
Because the scale is logarithmic, each whole number change in pH represents a tenfold change in hydrogen ion concentration. A solution with pH 3 is ten times more acidic than one with pH 4.
Example 16: pH Calculation
(a) Find the pH of a solution with $[\text{H}^+] = 3.2 \times 10^{-5}$ mol/L.
$\text{pH} = -\log(3.2 \times 10^{-5}) = -(\log 3.2 + \log 10^{-5}) = -(0.505 - 5) \approx 4.49$
(b) Find $[\text{H}^+]$ for a solution with pH $= 7.4$ (blood).
$[\text{H}^+] = 10^{-7.4} \approx 3.98 \times 10^{-8}$ mol/L.
6.7 Practice Problems
Work through the following problems to solidify your understanding of exponential and logarithmic functions. Each problem has a detailed solution that you can reveal after attempting it yourself.
Problem 1
Evaluate without a calculator: $\log_4 64$
Show Solution
We need to find $y$ such that $4^y = 64$.
Since $4^3 = 64$, we have $\log_4 64 = 3$.
$$\boxed{3}$$
Problem 2
Expand completely: $\ln\!\left(\dfrac{e^2 \cdot x^5}{y^3}\right)$
Show Solution
Apply the quotient rule, then the product rule, then the power rule:
$$\ln\!\left(\frac{e^2 \cdot x^5}{y^3}\right) = \ln(e^2 \cdot x^5) - \ln(y^3)$$
$$= \ln(e^2) + \ln(x^5) - \ln(y^3)$$
$$= 2 + 5\ln x - 3\ln y$$
$$\boxed{2 + 5\ln x - 3\ln y}$$
Problem 3
Solve: $3^{x+2} = 27^{x-1}$
Show Solution
Rewrite $27$ as $3^3$:
$$3^{x+2} = (3^3)^{x-1} = 3^{3x-3}$$
By the one-to-one property: $x + 2 = 3x - 3$
$$5 = 2x \implies x = \frac{5}{2}$$
$$\boxed{x = \frac{5}{2}}$$
Problem 4
Solve: $7^{3x} = 200$
Show Solution
Take the natural logarithm of both sides:
$$3x \cdot \ln 7 = \ln 200$$
$$x = \frac{\ln 200}{3\ln 7} = \frac{5.2983}{3(1.9459)} = \frac{5.2983}{5.8378} \approx 0.9076$$
$$\boxed{x \approx 0.9076}$$
Problem 5
Solve: $\log_2(x - 3) + \log_2(x + 1) = 5$
Show Solution
Use the product rule: $\log_2\big[(x - 3)(x + 1)\big] = 5$
Convert to exponential form: $(x - 3)(x + 1) = 2^5 = 32$
$$x^2 - 2x - 3 = 32$$
$$x^2 - 2x - 35 = 0$$
$$(x - 7)(x + 5) = 0$$
$x = 7$ or $x = -5$
Check: For $x = -5$: $\log_2(-8)$ is undefined. Reject.
For $x = 7$: $\log_2(4) + \log_2(8) = 2 + 3 = 5$. Valid.
$$\boxed{x = 7}$$
Problem 6
Write as a single logarithm: $3\log x - \frac{1}{2}\log(x + 4) + \log 5$
Show Solution
Apply the power rule, then product and quotient rules:
$$= \log x^3 - \log(x+4)^{1/2} + \log 5$$
$$= \log\!\left(\frac{5x^3}{\sqrt{x+4}}\right)$$
$$\boxed{\log\!\left(\dfrac{5x^3}{\sqrt{x + 4}}\right)}$$
Problem 7
You deposit $\$2{,}000$ in a savings account that pays $4.5\%$ annual interest, compounded monthly. How much will you have after 6 years?
Show Solution
Use $A = P\!\left(1 + \frac{r}{n}\right)^{nt}$ with $P = 2000$, $r = 0.045$, $n = 12$, $t = 6$:
$$A = 2000\!\left(1 + \frac{0.045}{12}\right)^{72}$$
$$= 2000(1.00375)^{72}$$
$$= 2000(1.30958\ldots)$$
$$\approx \$2{,}619.16$$
$$\boxed{\$2{,}619.16}$$
Problem 8
A population of deer grows according to $P(t) = 250 \cdot e^{0.12t}$, where $t$ is measured in years. (a) What is the initial population? (b) When will the population reach 1000?
Show Solution
(a) The initial population is $P(0) = 250 \cdot e^0 = 250$ deer.
(b) Solve $1000 = 250 \cdot e^{0.12t}$:
$$4 = e^{0.12t}$$
$$\ln 4 = 0.12t$$
$$t = \frac{\ln 4}{0.12} = \frac{1.3863}{0.12} \approx 11.55 \text{ years}$$
$$\boxed{(a)\ 250 \text{ deer} \qquad (b)\ \approx 11.55 \text{ years}}$$
Problem 9
A radioactive isotope has a half-life of 14 days. If a sample currently has 80 grams, how much will remain after 42 days?
Show Solution
Find $k$: $k = \dfrac{\ln 2}{14} \approx 0.04951$
Use $A(t) = 80 \cdot e^{-0.04951 \cdot 42}$:
$$A(42) = 80 \cdot e^{-2.0794} = 80 \cdot 0.125 = 10 \text{ grams}$$
Alternate method: 42 days = 3 half-lives. After each half-life, the amount halves:
$80 \to 40 \to 20 \to 10$ grams.
$$\boxed{10 \text{ grams}}$$
Problem 10
Solve: $2e^{3x} - 5e^{x} = 0$ (Hint: factor out $e^x$.)
Show Solution
Factor out $e^x$:
$$e^x(2e^{2x} - 5) = 0$$
Since $e^x > 0$ for all $x$, we need $2e^{2x} - 5 = 0$:
$$e^{2x} = \frac{5}{2}$$
$$2x = \ln\!\left(\frac{5}{2}\right)$$
$$x = \frac{1}{2}\ln\!\left(\frac{5}{2}\right) = \frac{\ln 5 - \ln 2}{2} \approx \frac{1.6094 - 0.6931}{2} \approx 0.4581$$
$$\boxed{x = \frac{1}{2}\ln\!\left(\frac{5}{2}\right) \approx 0.4581}$$
Problem 11
The pH of lemon juice is approximately 2.3. What is the hydrogen ion concentration $[\text{H}^+]$?
Show Solution
Use $\text{pH} = -\log[\text{H}^+]$:
$$2.3 = -\log[\text{H}^+]$$
$$\log[\text{H}^+] = -2.3$$
$$[\text{H}^+] = 10^{-2.3} \approx 5.01 \times 10^{-3} \text{ mol/L}$$
$$\boxed{[\text{H}^+] \approx 5.01 \times 10^{-3} \text{ mol/L}}$$
Problem 12
How long will it take $\$10{,}000$ to grow to $\$25{,}000$ at $5\%$ annual interest, compounded continuously?
Show Solution
Use $A = Pe^{rt}$ with $A = 25000$, $P = 10000$, $r = 0.05$:
$$25000 = 10000 \cdot e^{0.05t}$$
$$2.5 = e^{0.05t}$$
$$\ln 2.5 = 0.05t$$
$$t = \frac{\ln 2.5}{0.05} = \frac{0.9163}{0.05} \approx 18.33 \text{ years}$$
$$\boxed{\approx 18.33 \text{ years}}$$