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Week 14 · Chapter

Chapter 14 — Logarithmic Functions

College Algebra Free sample edition

The module's primary reading · ~40–50 minutes · Objective 8 (analyze exponential and logarithmic functions and their graphs, apply the properties of logarithms, and solve exponential and logarithmic equations and applied growth, decay, and compound-interest problems), Objective 7 (analyze polynomial and rational functions (end behavior, zeros and multiplicity, domain, and asymptotes), and simplify and solve rational and radical expressions and equations, including expressions with rational exponents, checking for extraneous solutions) · Read it first this week; the lecture, the tutorial, the quiz, and the Model/Build Studio all build on it.

In plain language

This week you learn to ask a power for its exponent. Last week you fed an input into a growth rule and got an output; now you know the output and want the input back. A logarithm (the exponent you need to turn a base into a given number) answers that question. You will switch between a logarithm and the power it describes, work out logarithms in your head, and draw the graph of a logarithmic function (a rule whose output is a logarithm). Then you will use three rules that let you split one logarithm into simpler pieces or merge several into one, and you will learn the one move that has no rule at all.

Why you already care about this

A science lab keeps a 100-gram sample that loses half its mass every period. After 1 period it holds 50 grams, after 2 it holds 25 grams, and after 3 it holds 12.5 grams. Now flip the question around. Someone hands you the jar, the label says 12.5 grams, and you want to know how many periods have passed. You are asking for an exponent: how many times was 100 cut in half?

That same flipped question shows up all over. A culture of 5 cells that doubles every hour reaches 40 cells after some number of hours; how many? A video shared by 1 person, then 2, then 4, reaches 1024 shares after some number of doublings; how many? A sound meter, a pH strip, and an earthquake report all use scales built on powers of 10, where equal steps on the scale stand for multiplying by the same amount. Each of these asks how many times something was multiplied. A logarithm is the tool built for exactly that question.

This chapter shows you how to read a logarithm, compute it, graph it, and rewrite it.

By the end of this chapter you will be able to:

  • answer an exponent question such as 3 · 2ˣ = 24 by reading a power backwards, and name the answer as a logarithm;
  • convert a statement between logarithmic form and exponential form in both directions;
  • evaluate logarithms in your head, including the log of 1, the log of the base, and negative logs of reciprocals;
  • use the common logarithm log(x) and the natural logarithm ln(x);
  • graph y = log₂(x) as the inverse of y = 2ˣ, and find the domain and vertical asymptote of a shifted logarithm;
  • expand and condense logarithms with the product, quotient, and power properties, and show with numbers why the log of a sum has no rule.

Carry this sentence with you: count the multiplications; a log is never a factor.

Where we are

  • Already covered: Week 1 set the exponent rules and the powers of two; Week 2 solved linear inequalities; Week 3 defined a function's domain; Week 10 named the vertical asymptote; Week 13 built exponential functions such as 3 · 2ˣ.
  • This week: the logarithm as an exponent, converting between forms, the common and natural logarithms, the graph and domain of a logarithmic function, and the three log properties.
  • How they connect: a logarithm runs a Week 13 exponential backwards, and each log property is a Week 1 exponent rule read through a logarithm.
  • Where we're going next: Week 15, Exponential & Logarithmic Equations & Applications, uses this week's conversions and properties to solve equations for an unknown exponent and to find doubling times and half-lives.

Each earlier week supplies a tool you use here. Week 1 listed the powers of two, 2⁰ = 1 up to 2¹⁰ = 1024, and gave you the product rule: when you multiply powers of the same base, you add the exponents. Week 1 also showed that a negative exponent asks for a reciprocal, so 2⁻² = 1 ⁄ 4. Week 2 taught you to solve an inequality such as x − 2 > 0 and to write the answer in interval notation. Week 3 said a domain is the set of inputs a rule accepts. Week 10 described a vertical asymptote as a line a graph approaches but never touches. Week 13 introduced 3 · 2ˣ and the natural base e, the constant e ≈ 2.71828. Every one of those returns this week.

1. Ask the exponent question

You undo operations all the time. Subtracting undoes adding. Halving undoes doubling. Last week you met a new operation, raising a base to a variable exponent, and this week you meet the move that undoes it.

Start with last week's exponential, f(x) = 3 · 2ˣ. Feed it x = 3 and you get 3 · 8 = 24. Now run it backwards. Suppose you know the output is 24 and you want the input.

Worked example — 3 · 2ˣ read backwards. Solve 3 · 2ˣ = 24 for x.

  1. Peel off the 3. The 3 multiplies the power, so divide both sides by 3: 2ˣ = 24 ⁄ 3 = 8.
  2. Ask the exponent question. Which power of 2 is 8? Check the powers of two: 2³ = 2 · 2 · 2 = 8.
  3. Answer. x = 3.
  4. Check by substitution. 3 · 2³ = 3 · 8 = 24. It matches.

In words: the input that makes 3 · 2ˣ equal 24 is 3.

Step 2 asked a question that needs its own name, because you will ask it again and again. "The exponent that turns 2 into 8" is written log₂(8), and it equals 3.

  • A logarithm is the exponent you need: log_b(x) = y means bʸ = x. So log₂(8) = 3 because 2³ = 8. The base b is positive and not 1, and x must be positive.

You read log₂(8) = 3 aloud as "log base 2 of 8 is 3." The small 2 is the base. The 8 in the parentheses is the number you are trying to reach.

Worked example — a culture that doubles. A lab culture starts with 5 cells and doubles every hour, so after t hours it holds 5 · 2ᵗ cells. When does it hold 40 cells?

  1. Set up. 5 · 2ᵗ = 40.
  2. Peel off the 5. Divide both sides by 5: 2ᵗ = 40 ⁄ 5 = 8.
  3. Ask the exponent question. t = log₂(8) = 3.
  4. Check. 5 · 2³ = 5 · 8 = 40.

In words: the culture reaches 40 cells after 3 hours.

Misconception check — the 3 that sneaks into the base. Students often treat 3 · 2ˣ = 24 as 6ˣ = 24, multiplying the 3 into the base before they take the exponent. It is tempting because 3 · 2 = 6 sits right there. The cure: the exponent belongs to the 2 alone, so the 3 must come off by division first. Test it: 3 · 2³ = 24, but 6³ = 216. Divide by the 3, then ask about the base.

Your memory hook for the week: the log is the exponent you need.

2. Turn log form into exponential form and back

Think about a recipe card and a shopping list. They describe the same meal in two layouts. Exponential form and logarithmic form are two layouts of the same fact.

  • Exponential form is a statement written with a power, such as 4³ = 64.
  • Logarithmic form is a statement written with a logarithm, such as log₄(64) = 3. It names an exponent as the output.

The definition is the bridge between them: log_b(x) = y means bʸ = x. Two things never change when you cross that bridge. The base of a logarithm, the b, stays the base. The log's value, the y, is always the exponent.

Worked example — log form to exponential form. Rewrite log₄(64) = 3.

  1. Find the base. The base is 4.
  2. Find the exponent. The log's value is 3, so 3 is the exponent.
  3. Find the result. The number inside, 64, is what the power produces.
  4. Write it. 4³ = 64.
  5. Check the arithmetic. 4 · 4 · 4 = 16 · 4 = 64.

Worked example — exponential form to log form. Rewrite 10² = 100 and 5⁰ = 1.

  1. First statement. The base is 10, the exponent is 2, and the result is 100. So log₁₀(100) = 2, which you will soon write as log(100) = 2.
  2. Second statement. The base is 5, the exponent is 0, and the result is 1. So log₅(1) = 0.

In words: the exponent always moves to the far side of the equal sign in log form, by itself.

Misconception check — the swapped roles. Students often write log₄(64) = 3 as 3⁴ = 64, putting the log's value in the base. It is tempting because the 3 stands alone and looks like the "main" number. The cure: say the definition out loud, "base to the answer gives the inside." Then 4 to the 3 gives 64. Test the swap: 3⁴ = 81, not 64, so the swap fails its own check.

3. Read a logarithm in your head

Most of the logs you will meet this week can be found without a calculator. The trick is to know a few powers well. Here the powers of two do all the work.

Worked example — the powers of two as a log table. Read each power of two backwards.

The powers of two from 2⁰ to 2¹⁰ and the base-2 logarithm of each value
Power Value Base-2 log of the value
2⁰ 1 0
2¹ 2 1
2² 4 2
2³ 8 3
2⁴ 16 4
2⁵ 32 5
2⁶ 64 6
2⁷ 128 7
2⁸ 256 8
2⁹ 512 9
2¹⁰ 1024 10
  1. A small one. 2³ = 8, so log₂(8) = 3.
  2. A middle one. 2⁶ = 64, so log₂(64) = 6.
  3. The big one. 2¹⁰ = 1024, so log₂(1024) = 10.
  4. The smallest one. 2⁰ = 1, so log₂(1) = 0.

In words: the second column holds the numbers you put into the log, and the last column holds the answers you get out. A video shared by 1 person that doubles its shares each hour reaches 1024 shares after log₂(1024) = 10 hours.

The table stops at 2⁰ = 1. What happens below 1? The negative-exponent rule from Week 1 answers it.

Worked example — a negative log. Find log₂(1⁄4).

  1. Ask the exponent question. Which power of 2 gives 1⁄4?
  2. Use a reciprocal. 2⁻² = 1 ⁄ 2² = 1 ⁄ 4.
  3. Answer. log₂(1⁄4) = −2.

In words: the log is negative even though 1⁄4 is a positive number. A negative log means "a reciprocal of a power," not "a negative number inside."

Two values come out the same in every base, and they save a lot of time:

  • b⁰ = 1, so log_b(1) = 0. For example, log₅(1) = 0.
  • b¹ = b, so log_b(b) = 1. For example, log₂(2) = 1.

Worked example — the halving sample. A 100-gram sample halves each period, so after t half-lives it holds A = 100 · (1⁄2)ᵗ grams. How many half-lives until it holds 12.5 grams?

  1. Set up. 100 · (1⁄2)ᵗ = 12.5.
  2. Peel off the 100. Divide both sides by 100: (1⁄2)ᵗ = 0.125 = 1⁄8.
  3. Flip to a power of two. (1⁄2)ᵗ = 1⁄8 means 2ᵗ = 8.
  4. Ask the exponent question. t = log₂(8) = 3 half-lives.
  5. Check by halving. 100 → 50 → 25 → 12.5 grams. That is three halvings.

In words: the jar labelled 12.5 grams has been through 3 half-lives.

Misconception check — the log of 1. Students often say log₅(1) = 1, or that it does not exist, because "nothing" seems to happen. The cure: ask the exponent question. Which power of 5 is 1? Any nonzero base to the 0 power is 1, so log₅(1) = 0. The log of 1 is 0 in every base, and the log of the base itself is 1.

4. Use the common and natural logarithms

Two bases show up so often that they get their own short names. One comes from our base-10 way of counting. The other comes from continuous growth, the natural base e you met last week.

  • The common logarithm is the logarithm with base 10. It is written log(x), with no base shown. So log(1000) = 3, because 10³ = 1000.
  • The natural logarithm is the logarithm with base e, the constant e ≈ 2.71828. It is written ln(x). So ln(e) = 1, because e¹ = e, and ln(1) = 0, because e⁰ = 1.

Other courses may write the common logarithm as log₁₀(x). It means the same thing.

Worked example — two common logs. Find log₁₀(1000) and log₁₀(100).

  1. First one. Which power of 10 is 1000? 10³ = 10 · 10 · 10 = 1000, so log₁₀(1000) = 3.
  2. Second one. Which power of 10 is 100? 10² = 100, so log₁₀(100) = 2.

In words: for a power of 10, the common log counts the zeros after the 1.

Worked example — steps on a base-10 scale. Some measuring scales are built on a logarithm of a ratio: you compare a reading to a fixed reference, and the scale reports the log of that comparison. The decibel scale for sound and the pH scale for acidity are built this way. Suppose a base-10 scale compares two readings to the same reference. The first is 1000 times the reference, and the second is 100 times the reference.

  1. First reading. log(1000) = 3, so it sits 3 steps up.
  2. Second reading. log(100) = 2, so it sits 2 steps up.
  3. Compare. 1000 ⁄ 100 = 10. The readings differ by 1 step: 1000 versus 100 is a 10-fold change.

In words: on a base-10 scale, one step up means ten times as much, not one more unit. That is why a log scale fits huge ranges on a short dial.

In practice (your tools). Open the free Desmos graphing calculator. Type log(1000) on one line and ln(e) on the next; Desmos shows 3 and 1. For a base other than 10 or e, type log_2(1024): Desmos sets the 2 as a small base and shows 10.

Misconception check — log(100) = 10. Students often say log₁₀(100) = 10 because 100 ⁄ 10 = 10, treating the log as division by the base. It is tempting because both questions involve 100 and 10. The cure: a log asks for an exponent, not a quotient. Ask which power of 10 makes 100: 10² = 100, so log₁₀(100) = 2. Test the slip: 10¹⁰ is ten billion, nowhere near 100.

5. Draw the logarithm as the inverse of the exponential

You already know how to undo an operation in words. Now you will see it as a picture. Picture a coat check: you hand over a coat and get a ticket, and later the ticket brings back the coat. The ticket counter runs the coat counter in reverse.

  • An inverse function is a function that undoes another one by swapping its inputs and outputs. The function y = log₂(x) undoes y = 2ˣ.
  • A logarithmic function is a function of the form f(x) = log_b(x). Its domain is the positive numbers, and its range is all real numbers.

Worked example — swap the points. Build the graph of y = log₂(x) from the graph of y = 2ˣ.

  1. List points on the exponential. y = 2ˣ passes through (−2, 1⁄4), (0, 1), (1, 2), and (3, 8), because 2⁻² = 1⁄4, 2⁰ = 1, 2¹ = 2, and 2³ = 8.
  2. Swap each pair. (−2, 1⁄4) becomes (1⁄4, −2). (0, 1) becomes (1, 0). (1, 2) becomes (2, 1). (3, 8) becomes (8, 3).
  3. Check one swap with a log. The point (8, 3) says log₂(8) = 3. That matches the table.
  4. Read the new graph. y = log₂(x) passes through (1⁄4, −2), (1, 0), (2, 1), and (8, 3).

In words: every point on the log graph is a point of the exponential with its two numbers traded.

Now describe the whole graph of y = log₂(x), feature by feature:

  • Domain (0, ∞). You can only take the log of a positive number, because 2 to any power is positive.
  • Range: all real numbers. Inputs near 0 give large negative outputs, and big inputs give big outputs.
  • x-intercept (1, 0). log₂(1) = 0.
  • Vertical asymptote x = 0. The graph hugs the y-axis as x shrinks toward 0, dropping lower and lower, but it never touches or crosses it.
  • Slow growth. From x = 1 to x = 1024 the output climbs only from 0 to 10.

The exponential y = 2ˣ has the horizontal asymptote y = 0. Its inverse has the vertical asymptote x = 0. The line y = x acts as a mirror between them: fold the page along that line and one graph lands on the other.

In practice (your tools). In Desmos, type y = 2^x on line 1, y = log_2(x) on line 2, and y = x on line 3. Click the point where line 2 crosses the x-axis; Desmos labels it (1, 0). Then click on line 1 at x = 3 and on line 2 at x = 8 to see (3, 8) and (8, 3) side by side.

Misconception check — the graph that crosses its asymptote. Students often sketch y = log₂(x) bending past the y-axis into negative inputs, as if it kept going left. It is tempting because most graphs you have drawn run across the whole page. The cure: ask whether log₂(−1) has a value. It would need 2ʸ = −1, and no power of 2 is negative. So nothing exists to the left of x = 0, and the curve only approaches the y-axis.

6. Find a domain and a vertical asymptote

A logarithm accepts only positive arguments. The argument of a logarithm is the number or expression inside its parentheses, such as the 8 in log₂(8). When the argument is an expression with x in it, the domain rule turns into a Week 2 inequality.

Worked example — a shifted logarithm. Find the domain and vertical asymptote of log₃(x − 2).

  1. Require a positive argument. x − 2 > 0.
  2. Solve. Add 2 to both sides: x > 2.
  3. Write the domain. (2, ∞).
  4. Find the asymptote. The argument is zero at x = 2, so the vertical asymptote is x = 2.
  5. Check with two points. At x = 3, log₃(3 − 2) = log₃(1) = 0, so the x-intercept is (3, 0). At x = 11, log₃(11 − 2) = log₃(9) = 2, because 3² = 9.

In words: the graph of log₃(x − 2) lives to the right of x = 2 and hugs that line without touching it.

Worked example — two more domains. Find the domain and the vertical asymptote of log(x + 5) and of ln(2x − 1).

  1. log(x + 5). Require x + 5 > 0. Subtract 5 from both sides: x > −5. The domain is (−5, ∞), and the vertical asymptote is x = −5.
  2. ln(2x − 1). Require 2x − 1 > 0. Add 1 to both sides: 2x > 1. Divide both sides by 2: x > 1⁄2. The domain is (1⁄2, ∞), and the vertical asymptote is x = 1⁄2.

In words: set the argument greater than zero, solve, and the boundary you find is the asymptote.

Worked example — no real value. Evaluate log₂(−8).

  1. Ask the exponent question. Which y gives 2ʸ = −8?
  2. Look for one. Every power of 2 is positive. Even 2⁻² = 1⁄4 is positive.
  3. Answer. log₂(−8) has no real value.

Misconception check — log₂(−8) = −3. Students often answer log₂(−8) = −3, because log₂(8) = 3 and a minus sign seems to carry through. It is tempting because the numbers line up so neatly. The cure: test the claim in exponential form. 2⁻³ = 1⁄8, which is positive and nowhere near −8. A negative argument has no real logarithm, and a negative log comes from a reciprocal, not from a negative input.

An honesty note: the domain rule is a definition about real numbers. Some advanced courses extend logarithms to other kinds of numbers, and this course stays with the real numbers, where a negative argument has no log.

7. Use the three properties

Here is a pattern you have already seen. Multiply 8 by 4 and you get 32. In powers of two, that is 2³ · 2² = 2⁵. The exponents added: 3 + 2 = 5. Since a log is an exponent, multiplying inside a log should turn into adding outside it. It does.

The three properties of logarithms, for positive M and N:

  • The product property of logarithms: log_b(MN) = log_b(M) + log_b(N). The log of a product is the sum of the logs.
  • The quotient property of logarithms: log_b(M ⁄ N) = log_b(M) − log_b(N). The log of a quotient is the difference of the logs.
  • The power property of logarithms: log_b(Mᵖ) = p · log_b(M). The exponent inside the log comes down in front as a multiplier.

Worked example — check each property on the powers of two. Test all three rules with numbers from the log table.

  1. Product. log₂(8 · 4) = log₂(32) = 5. Separately, log₂(8) + log₂(4) = 3 + 2 = 5. The two sides agree.
  2. Quotient. log₂(64 ⁄ 4) = log₂(16) = 4. Separately, log₂(64) − log₂(4) = 6 − 2 = 4. They agree.
  3. Power. log₂(8²) = log₂(64) = 6. Separately, 2 · log₂(8) = 2 · 3 = 6. They agree.

In words: each property is a Week 1 exponent rule wearing a logarithm. Multiplying powers adds exponents, dividing subtracts them, and raising a power to a power multiplies them.

Misconception check — the power property used as a product. Students often turn log₂(8²) into log₂(8) · log₂(8), squaring the log instead of doubling it. It is tempting because the square seems to belong to the whole log. The cure: the exponent sits on the 8, inside the log, and it comes out as a multiplier. log₂(8) · log₂(8) = 3 · 3 = 9, but log₂(64) = 6. The check exposes the slip.

8. Expand and condense: no rule for a sum

Packing a suitcase and unpacking it use the same items in opposite directions. Expanding and condensing are the same two moves for logarithms.

  • To expand a logarithm is to rewrite one log of a product, quotient, or power as a sum or difference of simpler logs.
  • To condense logarithms is to combine a sum or difference of logs with the same base into a single log.

Worked example — expand log_b(x³y²). Write it as a sum of simpler logs.

  1. Split the product. The product property gives log_b(x³) + log_b(y²).
  2. Bring the exponents down. The power property gives 3 log_b(x) + 2 log_b(y).
  3. Check with numbers. Take b = 2, x = 2, y = 4. Then x³y² = 8 · 16 = 128, and log₂(128) = 7. The expanded side gives 3 · log₂(2) + 2 · log₂(4) = 3 · 1 + 2 · 2 = 3 + 4 = 7.

Worked example — expand ln(x⁴ ⁄ y³). Write it as a difference of simpler logs.

  1. Split the quotient. The quotient property gives ln(x⁴) − ln(y³).
  2. Bring the exponents down. The power property gives 4 ln(x) − 3 ln(y).

Worked example — condense 2 log(x) + log(y). Write it as one common log.

  1. Power property first. 2 log(x) becomes log(x²), so you have log(x²) + log(y).
  2. Then the product property. log(x²) + log(y) = log(x²y).
  3. Check with numbers. Take x = 10 and y = 10. The left side is 2 log(10) + log(10) = 2 + 1 = 3. The right side is log(10² · 10) = log(1000) = 3.

Worked example — condense 3 log(x) − log(y). Write it as one common log.

  1. Power property first. You get log(x³) − log(y).
  2. Then the quotient property. log(x³ ⁄ y).
  3. Check with numbers. Take x = 10 and y = 100. The left side is 3 log(10) − log(100) = 3 − 2 = 1. The right side is log(1000 ⁄ 100) = log(10) = 1.

Notice the order when you condense: coefficients go up into exponents first, and then the pieces merge.

Now the trap the whole week has been building toward. There is a property for the log of a product, a quotient, and a power. There is no property for the log of a sum.

Worked example — the sum trap. Compare log₂(4 + 4) with log₂(4) + log₂(4).

  1. Add first. 4 + 4 = 8, and log₂(8) = 3.
  2. Split the log. log₂(4) + log₂(4) = 2 + 2 = 4.
  3. Compare. 3 ≠ 4, so the log does not split over a sum.

In words: a log of a sum stays as it is. A sum of logs, read backwards through the product property, is the log of a product: log(x) + log(4) = log(4x), not log(x + 4). In base 2 at x = 4, log₂(4 · 4) = log₂(16) = 4, which matches 2 + 2 = 4.

Misconception check — log(x + 4) = log(x) + log(4). Students often split log(x + 4) into log(x) + log(4), because the product property looks like a "distribute the log" rule. It is tempting because distributing worked in Week 1. The cure: a log is not a number you multiply by, so it does not distribute. Test it with the sum trap: log₂(4 + 4) = 3, while log₂(4) + log₂(4) = 4. The sum of two logs belongs to a product, never to a sum.

In practice (your tools). In Desmos, type y = log(x + 4) on line 1 and y = log(x) + log(4) on line 2. Two different curves appear, so the split fails. Then change line 2 to y = log(4x) and compare it with a new line, y = log(x) + log(4): those two curves sit exactly on top of each other.

A caveat about checking with numbers: one matching value does not prove a rule, but one mismatch does disprove a claimed rule. That is why a single check, like 3 versus 4, is enough to reject the sum split.

A reflection to choose from. Pick one option, (a) or (b), and write two or three sentences.

  • (a) Think of a place in your own life or work where a quantity is multiplied over and over, such as a doubling, a halving, or a tenfold step. Describe the exponent question you would ask about it and the logarithm that answers it.
  • (b) Pick the trap from this chapter you are most likely to fall into: putting the log's value in the base, splitting a log over a sum, or giving a real value to the log of a negative number. Describe a short habit, such as a numeric check, that would catch it.

Key terms

  • Logarithm — The exponent you need: log_b(x) = y means bʸ = x, so log₂(8) = 3 because 2³ = 8; the base b is positive and not 1, and x must be positive. (first used: 1. Ask the exponent question)
  • Exponential form — A statement written with a power, such as 4³ = 64, which says the same thing as the logarithmic form log₄(64) = 3. (first used: 2. Turn log form into exponential form and back)
  • Logarithmic form — A statement written with a logarithm, such as log₄(64) = 3, which names an exponent as the output. (first used: 2. Turn log form into exponential form and back)
  • Base of a logarithm — The number b in log_b(x) that is raised to a power; it is the same base as in the matching exponential statement bʸ = x, and it must be positive and not equal to 1. (first used: 2. Turn log form into exponential form and back)
  • Common logarithm — The logarithm with base 10, written log(x) with no base shown, so log(1000) = 3 because 10³ = 1000. (first used: 4. Use the common and natural logarithms)
  • Natural logarithm — The logarithm with base e, the constant near 2.71828, written ln(x), so ln(e) = 1 and ln(1) = 0. (first used: 4. Use the common and natural logarithms)
  • Inverse function — A function that undoes another one by swapping its inputs and outputs, so y = log₂(x) undoes y = 2ˣ and the point (3, 8) on one graph becomes (8, 3) on the other. (first used: 5. Draw the logarithm as the inverse of the exponential)
  • Logarithmic function — A function of the form f(x) = log_b(x); its domain is the positive numbers, its range is all real numbers, and its graph has a vertical asymptote where the argument is zero. (first used: 5. Draw the logarithm as the inverse of the exponential)
  • Argument of a logarithm — The number or expression inside the parentheses of a logarithm, such as the 8 in log₂(8); it must be positive for the logarithm to have a real value. (first used: 6. Find a domain and a vertical asymptote)
  • Product property of logarithms — The rule log_b(MN) = log_b(M) + log_b(N) for positive M and N: the log of a product is the sum of the logs. (first used: 7. Use the three properties)
  • Quotient property of logarithms — The rule log_b(M ⁄ N) = log_b(M) − log_b(N) for positive M and N: the log of a quotient is the difference of the logs. (first used: 7. Use the three properties)
  • Power property of logarithms — The rule log_b(Mᵖ) = p · log_b(M) for positive M: the exponent inside the log comes down in front as a multiplier. (first used: 7. Use the three properties)
  • Expand a logarithm — To rewrite one logarithm of a product, quotient, or power as a sum or difference of simpler logarithms, as log_b(x³y²) becomes 3 log_b(x) + 2 log_b(y). (first used: 8. Expand and condense: no rule for a sum)
  • Condense logarithms — To combine a sum or difference of logarithms with the same base into a single logarithm, as 2 log(x) + log(y) becomes log(x²y). (first used: 8. Expand and condense: no rule for a sum)

Summary

  • A logarithm is an exponent: it answers "which power of the base gives this number?"
  • Exponential form and logarithmic form say the same thing; the base stays the base.
  • The log of 1 is 0 and the log of the base is 1, in every base.
  • A reciprocal of a power has a negative log.
  • The common log has base 10; the natural log has base e.
  • The log graph is the exponential graph with every point's coordinates swapped.
  • Set the argument greater than zero to find the domain; the boundary is the vertical asymptote.
  • A negative argument has no real logarithm.
  • Product becomes sum, quotient becomes difference, power becomes multiplier.
  • There is no rule for the log of a sum.
  • The halving sample's count of half-lives is a base-2 log: 12.5 grams means log₂(8) = 3.

Check your understanding

Work these before looking at the answers — they are ungraded, and the answers (with reasoning) are at the bottom of the page.

  1. Rewrite log₃(81) = 4 in exponential form and check it. Then evaluate log₅(1⁄25) in your head.
  2. Find the domain and the vertical asymptote of log₆(x − 7), and give one point on its graph.
  3. Expand log₃(a²b⁵ ⁄ c) and check your expansion at a = 3, b = 3, c = 9.
  4. A classmate writes log₂(16 + 16) = log₂(16) + log₂(16). Show with numbers whether that is right, and write the expression that log₂(16) + log₂(16) really equals.

Answers (worked reasoning — read after attempting)

  1. The base is 3, the exponent is 4, and the result is 81, so the exponential form is 3⁴ = 81. Check: 3 · 3 · 3 · 3 = 9 · 9 = 81. For log₅(1⁄25), ask which power of 5 gives 1⁄25. Since 5² = 25, the reciprocal is 5⁻² = 1 ⁄ 25, so log₅(1⁄25) = −2. Putting the 4 in the base (4³ = 64) is the swapped-roles slip.
  2. The argument must be positive: x − 7 > 0. Add 7 to both sides: x > 7. The domain is (7, ∞), and the vertical asymptote is x = 7, where the argument is zero. For a point, make the argument 1: at x = 8, log₆(8 − 7) = log₆(1) = 0, so (8, 0) is the x-intercept. Writing the domain as [7, ∞) would include x = 7, where the argument is 0 and the log has no value.
  3. The quotient property gives log₃(a²b⁵) − log₃(c). The product property splits the first piece: log₃(a²) + log₃(b⁵) − log₃(c). The power property brings the exponents down: 2 log₃(a) + 5 log₃(b) − log₃(c). Check at a = 3, b = 3, c = 9. Inside: a²b⁵ ⁄ c = 9 · 243 ⁄ 9 = 243 = 3⁵, so the left side is log₃(243) = 5. The expanded side is 2 · 1 + 5 · 1 − 2 = 2 + 5 − 2 = 5, using log₃(9) = 2. Both sides are 5.
  4. Add first: 16 + 16 = 32, and log₂(32) = 5. The split side is log₂(16) + log₂(16) = 4 + 4 = 8. Since 5 ≠ 8, the classmate's split is wrong; a log never splits over a sum. Read backwards through the product property, log₂(16) + log₂(16) = log₂(16 · 16) = log₂(256) = 8, which matches.

Next up in this module: the lecture, the slide deck, the Readings & Resources page, the lecture tutorial, the practice set, and the quiz — and next week the product, quotient, and power properties become the method for solving exponential and logarithmic equations.

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