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Week 15 · Chapter

Chapter 15 — Exponential & Logarithmic Equations & Applications

College Algebra Free sample edition

The module's primary reading · ~45–55 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 2 (solve linear equations and linear inequalities in one variable, including absolute-value equations and inequalities, and express solution sets in interval notation) · 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 solve for an unknown that sits up in an exponent. Some of these puzzles you can answer by writing both sides as powers of the same number. The rest need a logarithm (the exponent you need to turn a base into a given number), which brings the unknown down where you can reach it. You will also solve equations with the unknown inside a logarithm, and you will throw out any answer that asks for the logarithm of a negative number. Then you will answer the questions that make this topic worth learning: how long until money doubles, how long until it triples, and how long until a decaying sample is cut in half.

Why you already care about this

You put 1,000 dollars in a savings account that earns 6% a year, compounded continuously. You leave it alone. How many years until the balance reads 2,000 dollars? A biology lab grows a culture that starts at 500 cells and doubles every hour; the next experiment needs 4,000 cells. When is the culture ready? A lab sample loses mass continuously, and a technician needs to know when only half of it remains.

Each of those is the same question in a different coat. You know the starting amount, you know the target, and you know the rule. What you do not know is the time, and the time sits in the exponent. In Week 13 you put a number into an exponential rule and read the output. In Week 14 you learned that a logarithm names an exponent. This week you put those two ideas together and solve.

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

  • solve an exponential equation by rewriting both sides as powers of the same base;
  • isolate a power before you solve, so a starting amount or a coefficient is out of the way;
  • solve an exponential equation by taking a logarithm of both sides, and give the exact answer with a decimal check;
  • solve a logarithmic equation by converting it, or by condensing it first, and reject an extraneous candidate with the domain check;
  • find a doubling time, a tripling time, and a half-life, and explain why the starting amount never changes them.

Carry this sentence with you: match the base, else take the log; then check the domain.

Where we are

  • Already covered: Week 1 set the exponent rules; Weeks 2 and 3 gave the substitution check and the domain; Weeks 11 and 12 named extraneous solutions; Weeks 13 and 14 built exponentials and logarithms.
  • This week: exponential equations by matching bases and by taking logs, logarithmic equations with the domain check, and doubling time, tripling time, and half-life.
  • How they connect: every solve this week uses a Week 14 conversion or property to undo a Week 13 exponential, and every answer ends with a Week 2 check.
  • Where we're going next: Week 16 is the finals week. It brings no new material, only the cumulative final and its review bundle.

Here is what you bring with you. Week 1 listed the powers of two, from 2⁰ = 1 up to 2¹⁰ = 1024, and the power rule, (aᵐ)ⁿ = aᵐⁿ. Week 13 wrote continuous growth and decay as A = A₀ · e^(kt), where A₀ is the starting amount and k is the continuous rate. Week 14 said that log_b(x) = y means bʸ = x, and gave you the product property, log_b(MN) = log_b(M) + log_b(N), and the power property, log_b(Mᵖ) = p · log_b(M). It also said the argument of a logarithm must be positive. All of it returns this week, often in a single problem.

1. Match the bases

Picture two stacks of identical blocks. If the stacks are the same height and the blocks are the same size, the stacks hold the same number of blocks. Powers work that way. If two powers of the same base are equal, their exponents must be equal too.

  • An exponential equation is an equation with the variable in an exponent, such as 2ˣ = 16 or 2ˣ = 10. It is solved by matching bases or by taking a logarithm of both sides.
  • The same-base property says that if two powers with the same base are equal, their exponents are equal: when bᵐ = bⁿ for a base b > 0 with b ≠ 1, then m = n.

Worked example — 2ˣ = 16. Find x.

  1. Write 16 as a power of 2. The powers of two say 2⁴ = 2 · 2 · 2 · 2 = 16.
  2. Rewrite the equation. 2ˣ = 2⁴.
  3. Match the exponents. Same base, so x = 4.
  4. Check. 2⁴ = 16. It works.

In words: 2 must be used as a factor 4 times to make 16.

Worked example — 3^(x + 1) = 27. The exponent is a whole expression this time.

  1. Write 27 as a power of 3. 3³ = 3 · 3 · 3 = 27, so 3^(x + 1) = 3³.
  2. Match the exponents. x + 1 = 3.
  3. Solve the small equation. Subtract 1 from both sides: x = 2.
  4. Check. 3^(2 + 1) = 3³ = 27. It works.

Worked example — 9ˣ = 3^(x + 4). The two sides start with different bases.

  1. Find a shared base. 9 = 3², so 9ˣ = (3²)ˣ.
  2. Use the power rule. (3²)ˣ = 3^(2x), so the equation reads 3^(2x) = 3^(x + 4).
  3. Match the exponents. 2x = x + 4.
  4. Solve. Subtract x from both sides: x = 4.
  5. Check. 9⁴ = 6561, and 3^(4 + 4) = 3⁸ = 6561. Both sides match.

In words: 9 is a power of 3 in disguise, so once you write it that way, the exponents can be compared.

Misconception check — the exponent read as a multiplier. Students often solve 2ˣ = 16 by dividing 16 by 2, as if it said 2 · x = 16. That gives x = 8. It is tempting because 2 and x sit side by side, and side by side has meant "multiply" since Week 1. The cure: check the answer in the original equation. 2⁸ = 256, and 256 ≠ 16. The x is an exponent, not a factor, so you must ask how many 2s make 16, which is 4.

In practice (your tools). In Desmos, type y = 2^x on line 1 and y = 16 on line 2. Click the point where the curve meets the flat line. Desmos labels it (4, 16), which is your answer read off the graph.

2. Isolate the power before you match

A coefficient in front of a power is like a cover on a box. You take the cover off before you look inside. In an exponential equation, you divide away the starting amount first, and only then compare powers.

To isolate the power means to divide both sides by the number multiplying the power, so the power stands alone. You met this move in Week 14 with 3 · 2ˣ = 24.

Worked example — 3 · 2ˣ = 24. Find x.

  1. Divide both sides by 3. 2ˣ = 24 ⁄ 3 = 8.
  2. Match bases. 8 = 2³, so x = 3.
  3. Check. 3 · 2³ = 3 · 8 = 24. It works.

Worked example — the growing culture. A culture starts with 500 cells and doubles every hour, so after t hours it holds N = 500 · 2ᵗ cells. When does it reach 4,000 cells?

  1. Set up the equation. 500 · 2ᵗ = 4000.
  2. Isolate the power. Divide both sides by 500: 2ᵗ = 4000 ⁄ 500 = 8.
  3. Match bases. 8 = 2³, so t = 3.
  4. Check. 500 · 2³ = 500 · 8 = 4,000 cells.

In words: the culture reaches 4,000 cells after 3 hours. A lab planner who knows that number can decide when to start the next step.

Worked example — the savings balance, solved for the time. In Week 13, 1,000 dollars at 6% compounded once a year followed A = 1000 · (1.06)ᵗ. When does the balance reach 1,123.60 dollars?

  1. Set up the equation. 1000 · (1.06)ᵗ = 1123.60.
  2. Isolate the power. Divide by 1000, and the right side becomes the given constant 1.1236.
  3. Match bases. That constant is (1.06)², so (1.06)ᵗ = (1.06)² and t = 2.
  4. Check. 1000 · (1.06)² = 1,123.60 dollars.

In words: the balance reaches 1,123.60 dollars after 2 years, which matches the balance table from Week 13.

Misconception check — multiplying the coefficient into the base. Students often turn 500 · 2ᵗ = 4000 into 1000ᵗ = 4000 by multiplying 500 by 2 first. It is tempting because the two numbers sit next to each other. The cure: the exponent belongs to the 2 alone, so the order of operations does the power before the multiplication. Divide by 500 first. Test the shortcut: 1000³ is a billion, nowhere near 4,000.

3. Take a logarithm of both sides

Some targets are not neat powers. The same culture of 500 cells, doubling every hour, needs to reach 5,000 cells. Dividing by 500 leaves 2ᵗ = 10, and 10 is not a power of 2. You can still trap the answer: 2³ = 8 and 2⁴ = 16, so t is between 3 and 4. To find it exactly, you need a new move.

To take a logarithm of both sides means to apply the same logarithm, ln or log, to each side of an equation. Equal numbers have equal logarithms, so the equation stays true. Then the power property, log_b(Mᵖ) = p · log_b(M), brings the exponent down as a multiplier.

Worked example — 2ˣ = 10. Find x exactly, then check it with a decimal.

  1. Trap the answer first. 2³ = 8 and 2⁴ = 16, and 10 sits between them, so x is between 3 and 4.
  2. Take ln of both sides. ln(2ˣ) = ln(10).
  3. Bring the exponent down. The power property gives x · ln 2 = ln 10.
  4. Divide by ln 2. x = ln 10 ⁄ ln 2.
  5. Check with a decimal. A calculator gives ln 10 ⁄ ln 2 ≈ 3.3219, which sits between 3 and 4 as step 1 said. Then 2 to the power 3.3219 comes out at about 10.

In words: x = ln 10 ⁄ ln 2 exactly, about 3.3219. The exact form is the answer; the decimal is the check. For the culture, the same steps give t = ln 10 ⁄ ln 2 ≈ 3.32 hours, rounded to two places, and that lands between 3 hours (4,000 cells) and 4 hours (8,000 cells).

You may also see the answer written log₂(10). That is the same number, because log₂(10) is the exponent that turns 2 into 10. The change-of-base formula, log_b(x) = ln(x) ⁄ ln(b), is the bridge between the two forms. It lets a calculator with only ln and log keys work in any base: log₂(10) = ln 10 ⁄ ln 2.

Worked example — the same equation with common logs. Solve 2ˣ = 10 with log instead of ln.

  1. Take log of both sides. log(2ˣ) = log(10).
  2. Bring the exponent down. x · log 2 = log 10.
  3. Use log 10 = 1. x · log 2 = 1, so x = 1 ⁄ log 2.
  4. Check. A calculator shows the same 3.3219 for 1 ⁄ log 2. Either logarithm works, as long as you use it on both sides.

Misconception check — ln on one side, log on the other. Students often take ln of the left side and log of the right side. That gives x · ln 2 = log 10 = 1, so x = 1 ⁄ ln 2, which is about 1.44. It is tempting because both buttons say "log" in some form. The cure: whatever you do to one side of an equation, you do to the other, and "ln" and "log" are different operations. Test the wrong answer: 2^1.44 is about 2.7, not 10. Pick one kind of logarithm and use it on both sides.

In practice (your tools). On a calculator, type ln(10)/ln(2) as one entry rather than rounding each log first. In Desmos, type y = 2^x and y = 10, then click the crossing; Desmos shows x = 3.322, which agrees with your exact answer.

One honesty note about rounding: ln 2 and ln 10 never end, so any decimal you write is a rounded check. Keep the full calculator values until the very last step, and round only at the end.

4. Solve a logarithmic equation by converting it

A logarithm is a question in disguise: "what exponent do I need?" When an equation tells you the answer to that question, you can rewrite it as a power and read off the unknown.

  • A logarithmic equation is an equation with the variable inside the argument of a logarithm, such as log₂(x − 1) = 3. It is solved by condensing to one logarithm, converting to exponential form, and checking the domain.

The domain comes first. In Week 14 you learned that the argument of a logarithm must be positive. So before you solve, write down which inputs are allowed.

Worked example — log₂(x − 1) = 3. Find x.

  1. Domain first. The argument x − 1 must be positive: x − 1 > 0, so x > 1.
  2. Convert to exponential form. log₂(x − 1) = 3 means 2³ = x − 1.
  3. Evaluate the power. 2³ = 8, so 8 = x − 1.
  4. Solve. Add 1 to both sides: x = 9.
  5. Domain check. 9 > 1, so x = 9 is allowed.
  6. Check. log₂(9 − 1) = log₂(8) = 3, because 2³ = 8. It works.

In words: the only solution is x = 9.

Misconception check — solving the inside without converting. Students often see log₂(x − 1) = 3 and set x − 1 = 3, giving x = 4. It is tempting because the 3 looks like the value of the inside. The cure: the 3 is an exponent, not the argument. Convert first: base 2 to the power 3 gives the argument. Test x = 4: log₂(4 − 1) = log₂(3), and 2³ = 8, not 3, so log₂(3) is not 3.

5. Condense, convert, and run the domain check

Some logarithmic equations hold two logarithms. You cannot convert two logs at once, so you first merge them into one with the product property from Week 14. Then you convert. Then comes the step that makes this section different: the domain check.

  • The domain check is the step of testing each candidate in every original logarithm's argument and rejecting any candidate that makes an argument zero or negative.

You met extraneous solutions in Week 11 with rational equations and in Week 12 with radical equations: candidates produced by correct steps that still fail the original equation. Logarithmic equations create them too, because condensing can hide a negative argument inside a product.

Worked example — log(x) + log(x − 3) = 1. Find x. Remember that log with no base means base 10.

  1. Domain first. Both arguments must be positive: x > 0 and x − 3 > 0. The second condition is stricter, so x > 3.
  2. Condense. The product property gives log(x(x − 3)) = 1.
  3. Convert, base 10. x(x − 3) = 10¹ = 10.
  4. Multiply out. x² − 3x = 10.
  5. Set it to zero. Subtract 10 from both sides: x² − 3x − 10 = 0.
  6. Factor. You need two numbers with product −10 and sum −3. Those are −5 and 2, so (x − 5)(x + 2) = 0.
  7. Read the candidates. x = 5 or x = −2.
  8. Domain check on x = 5. 5 > 3, so it is allowed. Check: log(5) + log(5 − 3) = log(5) + log(2) = log(5 · 2) = log(10) = 1. It works.
  9. Domain check on x = −2. −2 is not greater than 3. In the original equation it asks for log(−2), and every power of 10 is positive, so log(−2) has no real value. Reject x = −2 as extraneous.

In words: the only solution is x = 5. The candidate x = −2 came from correct algebra, but it fails the original equation.

Misconception check — keeping both factored answers. Students often stop at step 7 and report x = 5 or x = −2. It is tempting because two factors usually mean two answers, as they did in Week 7. The cure: a logarithmic equation has a domain, and the quadratic does not know about it. Put each candidate back into every original argument. x = −2 gives log(−2), which has no real value, so it is extraneous.

In practice (your tools). In Desmos, type y = log(x) + log(x - 3) and y = 1 on two lines. The curve starts at the vertical line x = 3 and meets y = 1 once, at x = 5. There is no crossing at x = −2, because the curve does not exist to the left of x = 3. The graph shows the domain check in a picture.

6. Find a doubling time

Back to the savings account from the start of the chapter. Money that grows continuously follows A = A₀ · e^(kt), where k is the continuous rate: the constant k in A = A₀ · e^(kt), giving the growth (k > 0) or decay (k < 0) per unit of time. At 6% compounded continuously, k = 0.06.

  • The doubling time is the time it takes a growing amount to become twice as large. For continuous growth A = A₀ · e^(kt) with k > 0, it is ln 2 ⁄ k, whatever the starting amount.

Worked example — 1,000 dollars at 6%, continuously. When does the balance reach 2,000 dollars?

  1. Set up. A = 1000 · e^(0.06t), so 1000 · e^(0.06t) = 2000.
  2. Isolate the power. Divide by 1000: e^(0.06t) = 2.
  3. Take ln of both sides. ln(e^(0.06t)) = ln 2. Since ln(e) = 1, the left side is 0.06t, so 0.06t = ln 2.
  4. Divide by 0.06. With the given constant ln 2 ≈ 0.6931, t = ln 2 ⁄ 0.06 ≈ 0.6931 ⁄ 0.06 ≈ 11.55 years.
  5. Check the size. 11.55 is a sensible number of years for money to double at 6%: more than a decade, less than two.

In words: the balance doubles in about 11.55 years.

Now watch the starting amount. Suppose you deposit 5,000 dollars instead. Then 5000 · e^(0.06t) = 10000, and dividing by 5000 again leaves e^(0.06t) = 2. The same steps follow, and you land on the same 11.55 years. For any starting amount A₀, the equation A₀ · e^(kt) = 2A₀ becomes e^(kt) = 2 once you divide by A₀. That is why the doubling time is ln 2 ⁄ k and nothing else.

A higher rate means faster doubling. At 8% the doubling time is t = ln 2 ⁄ 0.08 ≈ 8.66 years, nearly three years sooner than at 6%.

The rule of 72. The rule of 72 is a mental estimate of a doubling time in years: 72 divided by the percent growth rate. At 6% it says 72 ⁄ 6 = 12 years, close to the exact 11.55 years. It is a rule of thumb, not a law; the exact value is ln 2 ⁄ k. The estimate is handy when you are comparing two savings offers in your head.

One caveat about every model in this section: a real savings rate can change from year to year, and a doubling time assumes the rate stays fixed the whole time.

Misconception check — a bigger deposit takes longer to double. Students often think 5,000 dollars needs more time to double than 1,000 dollars, because 5,000 more dollars sounds like a bigger job than 1,000 more. The cure: at a fixed rate, growth is a percentage of whatever is there, so a bigger balance earns more each year in the same proportion. In the algebra, A₀ cancels when you divide both sides by it. Both deposits double in about 11.55 years at 6%.

In practice (your tools). In Desmos, type y = 1000*e^(0.06x) and y = 2000 and click the crossing; it sits near x = 11.55. Then change 1000 to 5000 and 2000 to 10000. The crossing does not move.

7. Find a tripling time

Doubling is the famous question, but any multiple works the same way. To triple, the power has to reach 3 instead of 2.

  • The tripling time is the time it takes a growing amount to become three times as large. For continuous growth A = A₀ · e^(kt) with k > 0, it is ln 3 ⁄ k, whatever the starting amount.

Worked example — 2,000 dollars at 5%, continuously. When does the balance reach 6,000 dollars?

  1. Set up. A = 2000 · e^(0.05t), so 2000 · e^(0.05t) = 6000.
  2. Isolate the power. Divide by 2000: e^(0.05t) = 3.
  3. Take ln of both sides. 0.05t = ln 3.
  4. Divide by 0.05. With the given constant ln 3 ≈ 1.0986, t = ln 3 ⁄ 0.05 ≈ 1.0986 ⁄ 0.05 ≈ 21.97 years.

In words: the balance triples in about 21.97 years. Any other starting amount at 5% would triple in the same time, because the 2,000 divided out in step 2.

Misconception check — tripling takes 1.5 times as long as doubling. Students often guess that if doubling takes some number of years, tripling takes half again as long, since 3 is 1.5 times 2. It is tempting because it treats growth like a straight line. The cure: the times are ln 2 ⁄ k and ln 3 ⁄ k, so their ratio is ln 3 ⁄ ln 2, about 1.58, not 1.5. Exponential growth does not follow the straight-line pattern of Week 4.

8. Find a half-life

Decay runs the same machinery downhill. A lab sample loses mass continuously and follows A = A₀ · e^(−0.15t), with t in hours. Here the continuous rate is k = −0.15 per hour, and the negative sign is what makes it decay. In Week 13 you defined a half-life as the time it takes a decaying amount to fall to half its size. Now you can compute it from k.

Worked example — the half-life when k = −0.15 per hour. How long until half the sample is left?

  1. Set up. Half the starting amount is (1 ⁄ 2) · A₀, so A₀ · e^(−0.15t) = (1 ⁄ 2) · A₀.
  2. Isolate the power. Divide by A₀: e^(−0.15t) = 1 ⁄ 2. The starting amount is already gone.
  3. Take ln of both sides. −0.15t = ln(1 ⁄ 2). By the quotient property, ln(1 ⁄ 2) = ln 1 − ln 2 = 0 − ln 2 = −ln 2. So −0.15t = −ln 2.
  4. Divide by −0.15. With the given constant ln 2 ≈ 0.6931, t = ln 2 ⁄ 0.15 ≈ 0.6931 ⁄ 0.15 ≈ 4.62 hours.

In words: the half-life is about 4.62 hours. In formula form, the half-life is −ln 2 ⁄ k, and with k = −0.15 the two minus signs make a positive time.

Worked example — three half-lives. A 400-gram sample decays by the same rule. How long until 50 grams are left?

  1. Count the halvings. 400 → 200 → 100 → 50 grams. That is three halvings, because 50 ⁄ 400 = 1 ⁄ 8 = (1 ⁄ 2)³.
  2. Multiply. Three half-lives take 3 · 4.62 = 13.86 hours.
  3. Check another way. Solving 400 · e^(−0.15t) = 50 directly gives t = ln 8 ⁄ 0.15, which also rounds to 13.86 hours.

In words: the sample falls from 400 grams to 50 grams in about 13.86 hours.

The halving table from Week 13 reads backwards the same way. The 100-gram sample measured in half-lives follows A = 100 · (1 ⁄ 2)ᵗ, and it reaches 12.5 grams when (1 ⁄ 2)ᵗ = 1 ⁄ 8, at t = 3 half-lives: 100 → 50 → 25 → 12.5 grams.

Misconception check — a dropped minus sign. Students often write the decay as e^(0.15t), or divide by 0.15 when the equation says −0.15. The first mistake turns decay into growth; the second gives a negative time. It is tempting because the minus sign is small and easy to lose. The cure: keep k = −0.15 attached to its sign all the way through, and check that your answer is a positive time. ln(1 ⁄ 2) is negative and −0.15 is negative, so their quotient is positive.

In practice (your tools). In Desmos, type y = 400*e^(-0.15x) and then y = 200, y = 100, and y = 50 on three more lines. Click each crossing. The x-values land near 4.62, 9.24, and 13.86. Equal steps in time, each one halving the mass.

9. Look back: one habit, fifteen weeks

Think about the very first problems in this course: 2 + 3 · 4² = 50, and the powers of two from 2⁰ up to 2¹⁰ = 1024. Those exponent rules are the reason 9ˣ could be rewritten as 3^(2x) this week. The check by substitution you learned in Week 2 is the step that caught the trap in Section 1. The domain rules from Week 3 became the domain check of Section 5. The factoring of Week 6 and the quadratics of Week 7 solved x² − 3x − 10 = 0. Weeks 11 and 12 taught you that correct steps can still produce a false candidate.

So the course has not been a pile of separate tricks. It has been a handful of moves you can always check: simplify, solve, graph, factor, and model, and then substitute to confirm. This week used almost all of them in one place.

Misconception check — "I have to memorize every type." Students often finish a term believing every chapter was its own island of rules to memorize. It is tempting because each week had new vocabulary. The cure: look at what each solve this week actually did. It undid an operation, kept the equation balanced, and checked the answer. Those three habits are the course, and they will carry you through the final and the next math course you take.

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 doubling time or a half-life would help you decide something, such as a savings goal, a dose schedule, or a growing project. Describe the equation you would set up and how you would check the answer.
  • (b) Pick one move from this course that you now use without thinking, such as checking by substitution or checking a domain. Describe the week you learned it and one place it saved you from a wrong answer this week.

Key terms

  • Exponential equation — An equation with the variable in an exponent, such as 2ˣ = 16 or 2ˣ = 10; it is solved by matching bases or by taking a logarithm of both sides. (first used: 1. Match the bases)
  • Same-base property — If two powers with the same base are equal, their exponents are equal: when bᵐ = bⁿ for a base b > 0 with b ≠ 1, then m = n, so 2ˣ = 2⁴ gives x = 4. (first used: 1. Match the bases)
  • Take a logarithm of both sides — To apply the same logarithm, ln or log, to each side of an equation so the power property can bring a variable exponent down as a multiplier, as ln(2ˣ) = ln(10) becomes x · ln 2 = ln 10. (first used: 3. Take a logarithm of both sides)
  • Change-of-base formula — The rule log_b(x) = ln(x) ⁄ ln(b) for positive x and a base b > 0 with b ≠ 1, which lets a calculator with only ln or log keys evaluate any base, so log₂(10) = ln 10 ⁄ ln 2 ≈ 3.3219. (first used: 3. Take a logarithm of both sides)
  • Logarithmic equation — An equation with the variable inside the argument of a logarithm, such as log₂(x − 1) = 3; it is solved by condensing to one logarithm, converting to exponential form, and checking the domain. (first used: 4. Solve a logarithmic equation by converting it)
  • Domain check — The step of testing each candidate in every original logarithm's argument and rejecting any candidate that makes an argument zero or negative, as x = −2 is rejected in log(x) + log(x − 3) = 1. (first used: 5. Condense, convert, and run the domain check)
  • Continuous rate — The constant k in a continuous model A = A₀ · e^(kt), giving the growth (k > 0) or decay (k < 0) per unit of time, so 6% continuous growth has k = 0.06 and the decaying sample has k = −0.15 per hour. (first used: 6. Find a doubling time)
  • Doubling time — The time it takes a growing amount to become twice as large; for continuous growth A = A₀ · e^(kt) with k > 0 it is ln 2 ⁄ k, whatever the starting amount. (first used: 6. Find a doubling time)
  • Rule of 72 — A mental estimate of a doubling time in years: 72 divided by the percent growth rate, so 72 ⁄ 6 = 12 years at 6%, against the exact ln 2 ⁄ 0.06 ≈ 11.55 years. (first used: 6. Find a doubling time)
  • Tripling time — The time it takes a growing amount to become three times as large; for continuous growth A = A₀ · e^(kt) with k > 0 it is ln 3 ⁄ k, whatever the starting amount. (first used: 7. Find a tripling time)

Summary

  • If both sides are powers of the same base, set the exponents equal.
  • Divide away a coefficient before you match bases or take a log.
  • If the bases will not match, take the same log of both sides and bring the exponent down.
  • The exact answer keeps the logs, such as ln 10 ⁄ ln 2; the decimal is a check.
  • Convert a logarithmic equation to exponential form; condense first if there are two logs.
  • Write the domain before you solve, and reject any candidate that makes an argument zero or negative.
  • Doubling time is ln 2 ⁄ k, tripling time is ln 3 ⁄ k, and half-life is −ln 2 ⁄ k.
  • The starting amount cancels, so it never changes a doubling time or a half-life.
  • The rule of 72 is an estimate, not the exact value.

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. Solve 4ˣ = 2^(x + 3) by matching bases, and check your answer in the original equation.
  2. A used car's value follows V = 18000 · e^(−0.1t) dollars, with t in years. How many years until the car is worth half its price? Use ln 2 ≈ 0.6931 and round to two places.
  3. Solve log₃(2x + 1) = 2. State the domain first, and check your answer.
  4. Solve log₂(x) + log₂(x + 2) = 3. Check both candidates and say which one, if any, is extraneous.

Answers (worked reasoning — read after attempting)

  1. Write 4 as 2², so 4ˣ = (2²)ˣ = 2^(2x) by the power rule. Now 2^(2x) = 2^(x + 3), so the exponents match: 2x = x + 3. Subtract x from both sides: x = 3. Check: 4³ = 64, and 2^(3 + 3) = 2⁶ = 64. Both sides match. Comparing 4 with 2 directly, without a shared base, is the step that goes wrong.
  2. Half the price is (1 ⁄ 2) · 18000, so 18000 · e^(−0.1t) = (1 ⁄ 2) · 18000. Divide by 18000: e^(−0.1t) = 1 ⁄ 2. Take ln of both sides: −0.1t = −ln 2. Divide by −0.1: t = ln 2 ⁄ 0.1 ≈ 0.6931 ⁄ 0.1 ≈ 6.93 years. The 18,000 cancelled, so any car on this rule loses half its value in about 6.93 years, whatever it cost.
  3. The argument must be positive: 2x + 1 > 0, so 2x > −1 and x > −1 ⁄ 2. Convert: log₃(2x + 1) = 2 means 3² = 2x + 1, so 9 = 2x + 1. Subtract 1: 8 = 2x. Divide by 2: x = 4. Domain check: 4 > −1 ⁄ 2. Check: log₃(2 · 4 + 1) = log₃(9) = 2, because 3² = 9. Setting 2x + 1 = 2 instead would give x = 1 ⁄ 2, the "solve the inside" slip.
  4. Domain first: x > 0 and x + 2 > 0, so x > 0. Condense: log₂(x(x + 2)) = 3. Convert: x(x + 2) = 2³ = 8. Multiply out and set to zero: x² + 2x − 8 = 0. Two numbers with product −8 and sum 2 are 4 and −2, so (x + 4)(x − 2) = 0, and x = −4 or x = 2. Check x = 2: log₂(2) + log₂(4) = 1 + 2 = 3. It works. Check x = −4: it asks for log₂(−4), which has no real value, so x = −4 is extraneous. The only solution is x = 2.

Next up in this module: the Readings & Resources page, the lecture tutorial, the practice set, the quiz, and the discussion; next week brings the cumulative final, where every method returns.

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