math-2017:F.BF.5math-2017:F.LE.4
math-2017:F.BF.5 Understand the inverse relationship between exponents and logarithms and use this relationship to solve problems involving logarithms and exponents.
math-2017:F.LE.4 For exponential models, express as a logarithm the solution to ab^(ct) = d where a, c, and d are numbers and the base b is 2, 10, or e; evaluate the logarithm using technology.
Learning goal: Understand the inverse relationship between exponents and logarithms and use this relationship to solve problems involving logarithms and exponents. For exponential models, express as a logarithm the solution to an equation of the form ab^(ct) = d, where a, c and d are numbers and the base b is 2, 10 or e, and evaluate the logarithm using technology.
Scope note. F.BF.5 (the inverse relationship between exponents and logarithms) is marked (+) in Ohio's Learning Standards for mathematics, which Ohio uses to indicate a standard "for students who are planning on taking advanced courses". It is taught here in full, as extension content for students on that path.
The question a logarithm answers. An exponential expression bˣ answers the question "what do I get when I raise b to the power x?": 2⁵ = 32. A logarithm answers the reverse question, "to what power must I raise b to get this number?": log₂ 32 = 5, read "the logarithm base 2 of 32 is 5", because 2⁵ = 32. In general, for a positive base b ≠ 1 and a positive number N, log_b N = x means exactly that bˣ = N. The two statements are the same fact written in two directions, and translating between them is the whole skill. Since 10³ = 1000, log₁₀ 1000 = 3. Since 10⁻² = 0.01, log₁₀ 0.01 = −2. Since 2⁰ = 1, log₂ 1 = 0, and the same holds for every base: log_b 1 = 0. Since b¹ = b, log_b b = 1. A logarithm of a negative number or of zero is undefined, because a positive base raised to any real power is positive.
Three bases matter most. Base 2 is the base of doubling: populations that double, halving of a dose of medicine written as 2^(−t/8), and everything in computing. Base 10 is the common logarithm, written log₁₀ or often just log on a calculator, and it is the base of scientific notation and of decibel and pH scales. The number e ≈ 2.71828 is the base of continuous growth, and its logarithm is the natural logarithm, written ln: ln N = x means eˣ = N, so ln e = 1, ln 1 = 0 and ln 20 ≈ 2.9957 because e^2.9957 ≈ 20. Calculators and graphing tools have keys for log₁₀ and ln, and most have a log₂ key or a general log_b function; when a tool lacks one, use log₂ N = ln N / ln 2, a change of base that follows from the definition.
Exponents and logarithms undo each other. Because y = log_b x means x = bʸ, the functions bˣ and log_b x are inverses of each other. Composing them in either order returns the input: log_b(bˣ) = x for every real x, and b^(log_b x) = x for every positive x. So log₂(2⁷) = 7, 10^(log₁₀ 5) = 5 and ln(e^0.3) = 0.3. The graph of log_b x is the reflection of the graph of bˣ across the line y = x: since 2ˣ passes through (0, 1), (1, 2) and (3, 8), log₂ x passes through (1, 0), (2, 1) and (8, 3), and the vertical intercept of the exponential becomes the horizontal intercept of the logarithm. The domain of log_b x is x > 0, because those are the outputs of bˣ. This inverse relationship is the tool for solving: when the unknown is in the exponent, applying the logarithm of the same base to both sides brings it down most directly. (A logarithm of any other base also works, through the change of base, and when d/a is a known power of the base, such as 16 = 2⁴, equating exponents gives the same answer.)
Solving ab^(ct) = d. An exponential model has the form ab^(ct), where a is the starting amount, b is the base, and c scales the time. Solving ab^(ct) = d for t asks when the model reaches the value d. When a ≠ 0, c ≠ 0, the base b is positive and not 1, and d/a > 0, the procedure is always the same (if d/a ≤ 0 there is no solution, because a power of a positive base is always positive). First isolate the power by dividing both sides by a: b^(ct) = d/a. Second, rewrite in logarithmic form, using the inverse relationship: ct = log_b(d/a). Third, divide by c: t = log_b(d/a) / c. This expression is the exact solution; technology then evaluates it. Consider a colony of 500 bacteria modeled by N(t) = 500 · 2^(0.2t) after t hours, and ask when it reaches 8,000. Divide: 2^(0.2t) = 8000/500 = 16. Rewrite: 0.2t = log₂ 16. Since 2⁴ = 16, log₂ 16 = 4, so 0.2t = 4 and t = 20 hours. Check: 500 · 2^(0.2 · 20) = 500 · 2⁴ = 8000. When the logarithm is not a whole number the same steps produce an expression that only technology can turn into a decimal. Suppose a quantity grows as Q(t) = 3 · 10^(0.4t) and the question is when Q reaches 750. Divide: 10^(0.4t) = 250. Rewrite: 0.4t = log₁₀ 250. Divide: t = log₁₀ 250 / 0.4. A calculator gives log₁₀ 250 ≈ 2.3979, so t ≈ 2.3979/0.4 ≈ 5.995. The check 3 · 10^(0.4 · 5.995) ≈ 750.1 confirms it, the small excess coming from rounding t. With base e, a deposit growing as A(t) = 40e^(0.05t) reaches 100 when e^(0.05t) = 2.5, so 0.05t = ln 2.5 ≈ 0.9163 and t ≈ 0.9163/0.05 ≈ 18.33 years; the check 40e^(0.05 · 18.33) ≈ 100.0 confirms it. In each case the exact answer is a logarithm, and the decimal is an approximation of it, so write both: t = ln 2.5 / 0.05 ≈ 18.33.
When the base is not 2, 10 or e. A quantity that halves every 8 hours is modeled by 80(1/2)^(t/8), and the base 1/2 can be rewritten with base 2: (1/2)^(t/8) = 2^(−t/8). To find when 80 milligrams has fallen to 5 milligrams, solve 80 · 2^(−t/8) = 5. Divide: 2^(−t/8) = 5/80 = 1/16. Rewrite: −t/8 = log₂(1/16) = −4, since 2⁻⁴ = 1/16. So t = 32 hours, which makes sense because four halvings turn 80 into 40, 20, 10 and 5. A model with a base such as 1.05 can be handled with any logarithm key available by the same three steps, but this lesson keeps to the three bases named above, which cover doubling, powers of ten and continuous growth.
The misconception. Students sometimes treat 2^(0.2t) = 16 as if it were 2 · 0.2t = 16, multiplying the base by the exponent and then dividing both sides by the base. The exponent is not a factor, and none of the four arithmetic operations pulls it down; a logarithm does, because the logarithm is by definition the inverse of the exponential (unless d/a happens to be a known power of the base, when equating exponents also works). A second error is applying the logarithm before isolating the power, writing log₂(500 · 2^(0.2t)) = log₂ 8000 and then not knowing how to continue; that route can be finished with the product rule for logarithms, but dividing by a first is shorter and less error-prone.
Worked translation between forms and evaluation: Rewrite each statement in the other form and evaluate. 2ˣ = 8 is log₂ 8 = x, and since 2³ = 8, x = 3. log₁₀ x = −2 is 10⁻² = x, so x = 0.01. ln x = 2 is x = e² ≈ 7.389. log₂ 10 = x is 2ˣ = 10, which has no whole-number solution; technology gives log₂ 10 ≈ 3.3219, and the check 2^3.3219 ≈ 9.9998 shows the rounding. In each case the base of the logarithm is the base of the power, the logarithm's output is the exponent, and the logarithm's input is the power's value.
Worked solution of a base-10 model: A quantity of pollutant in a lake is modeled by P(t) = 6 · 10^(0.03t) metric tons after t years, and the question is when P reaches 90 metric tons. Isolate the power: 10^(0.03t) = 90/6 = 15. Rewrite as a logarithm: 0.03t = log₁₀ 15. Divide: t = log₁₀ 15 / 0.03. Technology gives log₁₀ 15 ≈ 1.1761, so t ≈ 1.1761/0.03 ≈ 39.2 years. Check: 0.03 · 39.2 = 1.176 and 10^1.176 ≈ 15.0, so 6 · 15.0 ≈ 90 metric tons. The exact solution is the logarithm expression; 39.2 is its decimal approximation.
Worked solution of a base-e model with a check by substitution: An investment grows continuously as A(t) = 1200e^(0.04t) dollars after t years. When does it reach 3,000 dollars? Divide by 1200: e^(0.04t) = 2.5. Rewrite: 0.04t = ln 2.5. Divide: t = ln 2.5 / 0.04. Technology gives ln 2.5 ≈ 0.9163, so t ≈ 22.91 years. Substituting back, 0.04 · 22.91 ≈ 0.9164 and e^0.9164 ≈ 2.5, so A ≈ 1200 · 2.5 = 3,000 dollars. Note that the same ln 2.5 appeared in the 40e^(0.05t) = 100 example with a different c; the ratio d/a decides the logarithm, and c decides how it is scaled.
Representation task: Make a two-column table with eight rows translating between exponential and logarithmic form: 2³ = 8, 10⁴ = 10000, e⁰ = 1, 2⁻³ = 1/8, 10^0.5 ≈ 3.162, e¹ = e, 2^(1/2) = √2 and one more of your own, writing beside each its logarithmic form such as log₂ 8 = 3. Then plot y = 2ˣ through (0, 1), (1, 2), (2, 4) and (3, 8) and y = log₂ x through (1, 0), (2, 1), (4, 2) and (8, 3) on the same axes with the line y = x, and write one sentence explaining why the second graph is the reflection of the first.
Independent performance: Write three exponential models of your own in the form ab^(ct), one with base 2, one with base 10 and one with base e, each attached to a situation (a doubling population, a quantity growing by powers of ten, a continuously growing account or continuously cooling object). For each, pose a question of the form "when does the quantity reach d?", solve it by isolating the power, rewriting as a logarithm and dividing by c, write the exact answer as a logarithm expression, evaluate it with technology to two decimal places, and check by substituting the decimal back into the model. Include one model whose base must first be rewritten, such as (1/2)^(t/5) = 2^(−t/5). Success criteria: each exact solution is written in the form log_b(d/a)/c before any decimal appears; the substitution check returns d to within rounding; the logarithm's base matches the model's base; and the interpretation of t carries units.
An adult may read the prompts aloud. Answer independently with words, objects, drawings or a demonstration before opening the answer guide. Award 1 point for a correct result or claim and 1 point for a matching explanation or demonstration; where a question asks only for a result (a corrected sentence, a spelling, a count), a fully correct result earns both points, and the answer guide shows what a complete response contains. Record prompting separately. The total (two points per question) is local practice feedback, not an Ohio proficiency level.
Explain what log₂ 32 means, evaluate it, and write the same fact in exponential form.
log₂ 32 asks to what power 2 must be raised to give 32. Since 2⁵ = 32, log₂ 32 = 5. The exponential form is 2⁵ = 32: the base of the logarithm is the base of the power, the logarithm's value is the exponent, and the number inside the logarithm is the value of the power.
Evaluate log₁₀ 0.01, ln e and log₂ 1 without technology, giving the reason for each.
log₁₀ 0.01 = −2 because 10⁻² = 1/100 = 0.01. ln e = 1 because e¹ = e, and log_b b = 1 for every base. log₂ 1 = 0 because 2⁰ = 1, and log_b 1 = 0 for every base. Each value is the exponent that produces the number inside the logarithm.
Explain why the functions 2ˣ and log₂ x are inverses, and use that to simplify log₂(2⁷) and 2^(log₂ 5).
y = log₂ x means the same as x = 2ʸ, so each function reverses the other: log₂ sends a power of 2 back to its exponent, and 2ˣ sends an exponent to its power. Therefore log₂(2⁷) = 7 and 2^(log₂ 5) = 5. Graphically, log₂ x is the reflection of 2ˣ across the line y = x, with (3, 8) on the exponential matching (8, 3) on the logarithm.
A bacteria colony is modeled by N(t) = 500 · 2^(0.2t) after t hours. Express as a logarithm the time when N reaches 8,000 and evaluate it.
Divide by 500: 2^(0.2t) = 16. Rewrite: 0.2t = log₂ 16. Divide by 0.2: t = log₂ 16 / 0.2. Since 2⁴ = 16, log₂ 16 = 4 and t = 4/0.2 = 20 hours. Check: 500 · 2^(0.2 · 20) = 500 · 16 = 8,000.
Solve 3 · 10^(0.4t) = 750. Write the exact solution as a logarithm and then evaluate it using technology to three decimal places.
Isolate the power: 10^(0.4t) = 250. Rewrite: 0.4t = log₁₀ 250. Divide: t = log₁₀ 250 / 0.4, the exact solution. Technology gives log₁₀ 250 ≈ 2.398, so t ≈ 5.995. Substituting back, 3 · 10^(0.4 · 5.995) ≈ 750, confirming the value.
An account grows as A(t) = 40e^(0.05t). Express as a logarithm the time when the balance reaches 100 and evaluate it.
Divide by 40: e^(0.05t) = 2.5. Rewrite with the natural logarithm: 0.05t = ln 2.5. Divide: t = ln 2.5 / 0.05. Technology gives ln 2.5 ≈ 0.9163, so t ≈ 18.33. The exact answer is the logarithm expression and 18.33 is its approximation; the check 40e^(0.05 · 18.33) ≈ 100 confirms it.
A dose of 80 milligrams halves every 8 hours, M(t) = 80(1/2)^(t/8). Rewrite the model with base 2 and find when the amount is 5 milligrams.
(1/2)^(t/8) = 2^(−t/8), so M(t) = 80 · 2^(−t/8). Set 80 · 2^(−t/8) = 5: divide to get 2^(−t/8) = 1/16, rewrite as −t/8 = log₂(1/16) = −4 because 2⁻⁴ = 1/16, and solve: t = 32 hours. Four halvings take 80 to 40, 20, 10 and then 5, which agrees.
A student solves 2^(0.2t) = 16 by writing 2 · 0.2t = 16 and getting t = 40. Explain the error and the correct reasoning.
The exponent 0.2t is not a factor of 2; 2^(0.2t) is 2 raised to the power 0.2t, not 2 times 0.2t. A logarithm brings the exponent down. The most direct choice is the logarithm of the same base, because log₂ is the inverse of 2ˣ: 0.2t = log₂ 16 = 4. Any other base works too, for example 0.2t · ln 2 = ln 16, and here one can also write 16 = 2⁴ and equate exponents, 0.2t = 4. Every correct route gives 0.2t = 4, so t = 20, and the check 2^(0.2 · 20) = 2⁴ = 16 works, while 2^(0.2 · 40) = 2⁸ = 256 does not.