UNIT 1Exponential functionsLesson 1 of 11 Exponential Growth and Decay

Exponential Growth and Decay

A1.5Algebra 1Functions20 practice questions

Some amounts grow or shrink by the same percent each step. They follow the rule y = a · bˣ.

What you'll be able to do

Find the start amount and the factor in y = a · bˣ.
Turn a percent rate into a factor, and a factor back into a percent.
Tell a linear table from an exponential table.
Solve growth, decay, half-life and interest problems.

The big idea

Multiply, don't add. A linear rule adds the same amount each step. An exponential rule multiplies by the same number each step.

The rule. In y = a · bˣ, a is the start amount, the value when x = 0. b is the factor you multiply by each step.

Growth or decay. If b is more than 1, the amount grows. If b is between 0 and 1, the amount shrinks.

Percent to factor. To grow 6%, you keep all of it and add 6%. So b = 1.06. To lose 15%, you keep 85%. So b = 0.85.

Tables. When x goes up by 1 each step, divide each y by the y before it. If you get the same number each time, the table is exponential and that ratio is b. If x goes up by 2 each step, that ratio is b², so take its square root to find b. If x = 0 is not in the table, divide backward by b to find the start amount a.

Never zero. Decay takes a part of what is left. The amount gets close to 0 but never gets there.

Interest. A bank adds a percent of what is in the account. The rate is r, paid n times a year, for t years. Then A = P(1 + r/n)ⁿᵗ.

Words you need

Exponential function: A rule y = a · bˣ. It multiplies by the same factor b each time x goes up by 1.
Start amount: The number a in y = a · bˣ. It is the value of y when x = 0.
Factor: The number you multiply by each step. It is b in y = a · bˣ.
Growth factor: A factor more than 1. A 6% rise has growth factor 1.06.
Decay factor: A factor between 0 and 1. A 15% drop has decay factor 0.85.
Percent rate: How much the amount goes up or down each step, as a percent.
Base: The number being raised to a power. In bˣ, b is the base.
Exponent: The small raised number. It says how many times to use the base as a factor.
Power: A base with an exponent, like 2⁵. It also means the exponent itself.
Doubling time: The time it takes an amount to double.
Half-life: The time it takes half of an amount to break down or go away.
Compound interest: Interest paid on the money in the account plus the interest it has already earned.
Principal: The money you put in at the start. It is P in the interest rule.
Square root: A number that gives the first number when you multiply it by itself. The square root of 2.25 is 1.5.
Asymptote: A line that a graph gets closer and closer to but never touches. For y = a · bˣ, the line y = 0 (the x-axis) is the asymptote.

Percent to factor

The story says	Factor b
Grows 6% each year	1.06
Loses 15% each year	0.85
Doubles	2
Halves	0.5
Grows 140%	2.4

Worked examples

Example 1

A phone costs $800. It loses 15% of its value each year. Write a rule for its value after x years. What is it worth after 3 years? Round to the nearest dollar.

It keeps 100% − 15% = 85% of its value each year. So b = 0.85.
It starts at $800. So a = 800, and the rule is y = 800(0.85)ˣ.
At x = 3, y = 800(0.85)³ = 800(0.614125) = 491.3. That rounds to $491.

Answer: y = 800(0.85)ˣ. It is worth about $491

Example 2

Is this table linear or exponential? x: 0, 1, 2, 3 and y: 5, 15, 45, 135. Write its rule.

The differences are 10, 30 and 90. They are not the same, so it is not linear.
The ratios are 15 ÷ 5 = 3, 45 ÷ 15 = 3 and 135 ÷ 45 = 3. They are the same, so it is exponential.
At x = 0, y = 5. So a = 5, b = 3, and the rule is y = 5(3)ˣ.

Answer: Exponential, y = 5(3)ˣ

Example 3

Graph y = 2(2)ˣ and y = 32(0.5)ˣ from x = 0 to x = 4. Where do the two graphs cross?

The growth rule gives 2, 4, 8, 16 and 32.
The decay rule gives 32, 16, 8, 4 and 2.
Both rules give 8 at x = 2. So the graphs cross at (2, 8).
The decay graph falls toward 0 but never touches it.

Answer: They cross at (2, 8)

Example 4

Maya puts $2,000 into an account that pays 4% annual interest, compounded quarterly. How much is in the account after 3 years? Round to the nearest cent.

Quarterly means n = 4 times a year. The rate per quarter is 0.04 ÷ 4 = 0.01, so the factor each quarter is 1.01.
In 3 years, there are 4 × 3 = 12 quarters.
A = 2,000(1.01)¹² ≈ 2,253.65.

Answer: $2,253.65

Common mistakes

Using the percent lost as the factor. A 15% loss keeps 85%. Use 0.85, not 0.15.

Adding the same amount each year. Growth of 20% on $1,000 is $200 only in year 1. After that, take 20% of the new amount.

Thinking +20% and −20% cancel. The second percent is taken from a new amount. 1.2 × 0.8 = 0.96, not 1.

Counting hours instead of periods. Doubling every 20 minutes for 2 hours is 6 doublings, not 2.

Using the yearly rate for monthly interest. Divide the rate by 12, and raise to 12 times the years.

Practice: 20 questions

15 of 20 are word problems. For a word problem, sketch it first and mark what you know.

Tier 1 Foundations Use the rule directly

A patient takes 400 milligrams of ibuprofen. The milligrams of medicine left in the body after x hours is modeled by y = 400(0.75)ˣ. What do 400 and 0.75 mean in this story, including units? How many milligrams are left after 2 hours? Word problem
Kenji buys a used car for $18,000. The car loses 8% of its value each year. He writes y = 18,000(0.8)ˣ for the car's value in dollars after x years. Find his mistake and write the correct equation. Word problem
Look at these four equations: A is y = 0.4(2.5)ˣ, B is y = 80(0.965)ˣ, C is y = 12(5/4)ˣ, and D is y = 300(0.82)ˣ. Which equations show exponential decay? Give the percent rate of change for all four equations, and state which decay equation loses a greater percent each step.
Three tables all use x: 0, 1, 2, 3. Table A has y: 120, 90, 60, 30. Table B has y: 160, 120, 90, 67.5. Table C has y: 160, 150, 130, 100. Decide whether each table is linear, exponential, or neither, explain why, and write the equation for the exponential table.
A cup of broth cools in a 0°C cold room. Its temperature y in °C after x minutes is modeled by y = 85(0.9)ˣ. Where does the graph of this rule cross the y-axis? Does the graph rise or fall from left to right? According to the equation, does the graph ever reach y = 0? Explain why. Word problem

Tier 2 Intermediate One more step to think through

A town had 12,000 people in 2020 and 12,720 people in 2021. The population keeps growing by the same percent each year. Write an equation for the population y at x years after 2020. How many people will the town have in 2025? Round to the nearest person. Word problem
A lab dish starts with 300 bacteria. The number of bacteria doubles every 20 minutes. Write an equation for the number of bacteria y after t minutes, and find how many bacteria are in the dish after 2 hours. Word problem
A winter coat costs $150. Its price is raised by 20% in October, then cut by 20% for a November sale. Amara says the coat is back to $150 because +20% and −20% cancel out. Find her mistake and give the actual sale price in dollars. Word problem
Mateo is offered two 6-day dog-sitting plans. Plan A pays $12 on Day 1, and adds $12 to the daily pay each day. Plan B pays $3 on Day 1, and doubles the daily pay each day. On which day does Plan B first pay more for that single day than Plan A? Which plan pays more in total across all 6 days, and by how much? Word problem
An exponential function has the table x: 0, 2, 4, 6 and y: 16, 36, 81, 182.25. Write the equation in the form y = a · bˣ, where b > 0. State the percent growth rate for each 1-unit step in x.

Tier 3 Advanced Several steps - find the hidden structure

Priya invests $10,000 for 4 years at a 6% annual interest rate. Bank A compounds interest once a year. Bank B compounds interest monthly. How much more money does Priya have after 4 years with Bank B than with Bank A? Round to the nearest cent. Word problem
A hospital stores a radioactive sample with a half-life of 8 days. After 24 days, 15 grams of the sample remain. How many grams did the sample have on Day 0, and how many grams will remain at Day 40? Word problem
Tariq puts $1,000 into a fund that grows by 20% each year. In which whole year does the fund first pass $2,400? Give the year and the fund's value in that year, rounded to the nearest cent. Word problem
Soren looks at the equation y = 100(1.25)⁻ˣ and says it shows exponential growth with a 25% growth rate because the number inside the parentheses is 1.25. Find his mistake, rewrite the equation in the form y = a · bˣ, and give the true percent rate of change.
The graph shows an exponential curve y = a · bˣ passing through (2, 144), (3, 108), and (4, 81). Find the starting amount a, the y-intercept at x = 0, and write the equation of the curve in the form y = a · bˣ.

Tier 4 Challenge Stretch problems, some above grade level

A new farm tractor loses 48.8% of its original value over its first 3 years, losing the same percent of its value each year. What percent of its value does the tractor lose each single year? Word problem
Elena compares two timber forests. Forest A grows in wood volume by 10% every year. Forest B grows in wood volume by 20% every 2 years. Which forest grows faster over a 2-year period, and by how many percentage points? Word problem
A filter removes half of the chlorine in a pool every 2 hours. Dev says the pool loses 25% of its chlorine each hour, so 0% of the chlorine is left after 4 hours. Find both of his mistakes. Then find what percent of the chlorine is actually removed in 1 hour, and what percent of the original chlorine remains after 4 hours. Round the hourly percent to the nearest tenth of a percent. Word problem
Pond A starts with 1,024 lily pads and increases by 25% each week. Pond B starts with 250 lily pads and doubles each week. In which week x do the two ponds have the exact same number of lily pads, and how many lily pads is that? Word problem
The logarithm log_b(n) is the power you raise b to in order to get n. For example, log₂(32) = 5 because 2⁵ = 32. A lab culture starts with 5 grams of yeast and triples in mass every hour, so its mass in grams after t hours is M = 5 · 3ᵗ. (a) Write the exact time t when the mass reaches 405 grams as a logarithm, and find its value in hours. (b) Without a calculator, between which two consecutive whole hours does the mass reach 250 grams? Explain how you know. Word problem Above grade level

Answers

Tier 1 Foundations

400 is the starting dose in milligrams at 0 hours; 0.75 is the decay factor, meaning 75% of the medicine remains (and 25% is broken down) each hour; 225 milligrams are left after 2 hours. At x = 0, y = 400(0.75)⁰ = 400 milligrams at the start. Multiplying by 0.75 each hour keeps 75% of the medicine and removes 25% per hour. At x = 2, y = 400(0.75)² = 400(0.5625) = 225 milligrams.
Kenji used 0.8 (a 20% drop) instead of subtracting 8% from 100%; the correct equation is y = 18,000(0.92)ˣ. An 8% loss means 100% − 8% = 92% of the value remains each year. Write 92% as the decimal decay factor 0.92 (whereas 0.8 is 80%, a 20% loss). The correct equation is y = 18,000(0.92)ˣ.
Equations B and D show decay. The percent rates are: A grows by 150%, B loses 3.5%, C grows by 25%, and D loses 18% each step. Equation D loses a greater percent than Equation B. Decay happens when the base b is between 0 and 1: B (b = 0.965) and D (b = 0.82) show decay, while A (b = 2.5) and C (b = 5/4 = 1.25) show growth. For growth, subtract 1 from b: A is 2.5 − 1 = 1.50 = 150% growth, and C is 1.25 − 1 = 0.25 = 25% growth. For decay, subtract b from 1: B loses 1 − 0.965 = 0.035 = 3.5%, and D loses 1 − 0.82 = 0.18 = 18%, so D loses a greater percent.
Table A is linear (subtracts 30 each step); Table B is exponential (multiplies by 0.75 each step) with equation y = 160(0.75)ˣ; Table C is neither (subtracts 10, then 20, then 30). In Table A, the differences are −30 each step (constant difference), so Table A is linear. In Table B, the ratios are 120 ÷ 160 = 0.75, 90 ÷ 120 = 0.75, and 67.5 ÷ 90 = 0.75 (constant ratio), so Table B is exponential with a = 160 and b = 0.75, giving y = 160(0.75)ˣ. In Table C, neither the differences (−10, −20, −30) nor the ratios are constant, so it is neither.
It crosses the y-axis at (0, 85); it falls from left to right; it gets closer and closer to y = 0 but never reaches it because multiplying 85 by 0.9 repeatedly always gives a positive number. At x = 0, y = 85(0.9)⁰ = 85, so the y-intercept is (0, 85). Since the base 0.9 is between 0 and 1, the temperature decreases, so the graph falls from left to right. Because 85 and 0.9ˣ are always positive, y gets closer to 0 as x grows but never touches 0.

Tier 2 Intermediate

y = 12,000(1.06)ˣ; 16,059 people in 2025. Find the yearly growth factor: 12,720 ÷ 12,000 = 1.06 (6% growth per year). With starting population a = 12,000 in 2020, the equation is y = 12,000(1.06)ˣ. The year 2025 is x = 5 years after 2020: y = 12,000(1.06)⁵ ≈ 16,058.71, which rounds to 16,059 people.
y = 300(2)^(t/20); 19,200 bacteria. In t minutes, there are t/20 doubling periods of 20 minutes each, so y = 300(2)^(t/20). Convert 2 hours into minutes: 2 × 60 = 120 minutes. That gives 120 ÷ 20 = 6 doubling periods. Multiply 300 by 2⁶: 300 × 64 = 19,200 bacteria.
The 20% discount is taken from the higher $180 October price, not from $150; the actual sale price is $144. A 20% increase multiplies the price by 1.20: $150 × 1.20 = $180. A 20% cut multiplies the new $180 price by 0.80: $180 × 0.80 = $144. Because 1.20 × 0.80 = 0.96, the final price is 96% of the original price ($6 less than $150).
Plan B first pays more for a single day on Day 6 ($96 vs. $72); however, Plan A pays $63 more in total across all 6 days ($252 vs. $189). Daily pay for Plan A (Days 1 to 6): $12, $24, $36, $48, $60, $72. Daily pay for Plan B (Days 1 to 6): $3, $6, $12, $24, $48, $96. Plan B first passes Plan A on Day 6 ($96 > $72). Total over 6 days: Plan A = 12 + 24 + 36 + 48 + 60 + 72 = $252; Plan B = 3 + 6 + 12 + 24 + 48 + 96 = $189. Plan A pays $252 − $189 = $63 more in total.
y = 16(1.5)ˣ; 50% growth per 1-unit step. From the table, the starting value at x = 0 is a = 16. From x = 0 to x = 2 is a 2-unit step, so b² = 36 ÷ 16 = 2.25. Take the positive square root: b = √2.25 = 1.5, giving y = 16(1.5)ˣ and a 50% growth rate per unit step.

Tier 3 Advanced

$80.12 more with Bank B ($12,704.89 vs. $12,624.77). Bank A (yearly): A = 10,000(1 + 0.06)⁴ = 10,000(1.06)⁴ ≈ $12,624.77. Bank B (monthly): A = 10,000(1 + 0.06/12)^(12 × 4) = 10,000(1.005)⁴⁸ ≈ $12,704.89. Subtract: $12,704.89 − $12,624.77 = $80.12.
120 grams at Day 0; 3.75 grams at Day 40. Twenty-four days is 24 ÷ 8 = 3 half-lives. Work backward 3 half-lives by doubling 3 times: 15 × 2³ = 15 × 8 = 120 grams at Day 0. Day 40 is 16 days (2 more half-lives) after Day 24: 15 × (1/2)² = 15 ÷ 4 = 3.75 grams.
Year 5, when the fund reaches $2,488.32. Write the model: y = 1,000(1.2)ˣ. Calculate whole years: Year 1 is $1,200; Year 2 is $1,440; Year 3 is $1,728; Year 4 is $2,073.60. In Year 5, y = 1,000(1.2)⁵ = $2,488.32, which is the first whole year above $2,400.
The negative exponent flips the base to its reciprocal 1/1.25 = 0.8; the equation is y = 100(0.8)ˣ, which is exponential decay of 20% per step. By exponent rules, (1.25)⁻ˣ = (1 / 1.25)ˣ = (4/5)ˣ = (0.8)ˣ. Rewrite the equation as y = 100(0.8)ˣ. Since 0.8 < 1, it is exponential decay, and 1 − 0.80 = 0.20 means a 20% decay rate per step.
a = 256; y = 256(0.75)ˣ. From (2, 144) to (3, 108), the decay factor is b = 108 ÷ 144 = 0.75 (check: 108 × 0.75 = 81 at x = 4). Work backward 2 steps from x = 2 to x = 0 by dividing by 0.75 twice: a = 144 ÷ (0.75)² = 144 ÷ 0.5625 = 256. The equation is y = 256(0.75)ˣ.

Tier 4 Challenge

20% per year. If 48.8% of the value is lost after 3 years, then 100% − 48.8% = 51.2% = 0.512 of the original value remains. Set b³ = 0.512 = 512/1,000 and take the cube root: b = 8/10 = 0.8. A yearly factor of 0.8 means the tractor loses 1 − 0.8 = 0.20 = 20% of its value each year.
Forest A grows faster by 1 percentage point over 2 years (21% growth vs. 20% growth). Over 2 years, Forest A multiplies its volume by (1.10)² = 1.21. A factor of 1.21 means 21% growth over 2 years, because the second year's 10% also grows on top of the first year's 10% gain. Comparing 21% to Forest B's 20%, Forest A grows faster by 1 percentage point.
Chlorine is cut in half every 2 hours (not subtracting 25% of the original per hour); about 29.3% is removed each hour, and 25% of the original chlorine remains after 4 hours. After 4 hours (two 2-hour half-lives), the fraction remaining is (0.5)² = 0.25 = 25%, not 0%. For a 1-hour decay factor b, two hours of decay gives b² = 0.5, so b = √0.5 ≈ 0.7071. The percent removed in 1 hour is 1 − 0.7071 = 0.2929, or about 29.3% (because 25% per hour would leave 0.75² = 56.25% after 2 hours, not 50%).
Week 3, when both ponds have 2,000 lily pads. Write both models: Pond A is y = 1,024(1.25)ˣ and Pond B is y = 250(2)ˣ. Calculate each week: Week 1: Pond A = 1,280; Pond B = 500. Week 2: Pond A = 1,600; Pond B = 1,000. Week 3: Pond A = 1,600 × 1.25 = 2,000; Pond B = 1,000 × 2 = 2,000. Both ponds have 2,000 lily pads in Week 3.
(a) t = log₃(81) = 4 hours; (b) between 3 and 4 hours, because 3ᵗ = 50 and 3³ = 27 < 50 < 81 = 3⁴. For 405 grams, set 5 · 3ᵗ = 405 and divide by 5 first: 3ᵗ = 81. By the definition of a logarithm, t = log₃(81) = 4 hours because 3⁴ = 81. For 250 grams, 5 · 3ᵗ = 250 gives 3ᵗ = 50 (so t = log₃(50)). Since 3³ = 27 and 3⁴ = 81, t is between 3 and 4 hours.

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