Linear vs exponential comparison

Learning goals

Distinguishes constant additive change from constant multiplicative change.
Compares linear and exponential representations in tables, graphs, and equations.
Explains long-run differences between the models.
Selects the model consistent with a given pattern or context.

Difficulty: 1 – 5

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Linear vs exponential comparison 1

The table shows a function. <table> <tr><th>$x$</th><td>$0$</td><td>$1$</td><td>$2$</td><td>$3$</td></tr> <tr><th>$y$</th><td>$5$</td><td>$8$</td><td>$11$</td><td>$14$</td></tr> </table> Is the function linear or exponential? Give one piece of numerical evidence.

Hints

- Compare consecutive outputs over equal changes in $x$. - Decide whether the repeated pattern is additive or multiplicative.

Solution

1. Consecutive outputs change by $8-5=3$, $11-8=3$, and $14-11=3$. 2. The constant difference over equal input steps shows that the function is linear.

Answer

Linear; the output has a constant difference of $3$ for each increase of $1$ in $x$.

Linear vs exponential comparison 2

The table shows a function. <table><tr><th>$x$</th><td>$0$</td><td>$1$</td><td>$2$</td><td>$3$</td></tr><tr><th>$y$</th><td>$3$</td><td>$6$</td><td>$12$</td><td>$24$</td></tr></table> Is the function linear or exponential? Give the numerical pattern that proves your answer.

Hints

- Compare neighboring outputs in more than one way. - Does an additive or multiplicative pattern stay the same from step to step?

Solution

1. The ratios of consecutive outputs are $\frac{6}{3}=2$, $\frac{12}{6}=2$, and $\frac{24}{12}=2$. 2. A constant ratio for equal input steps shows exponential behavior.

Answer

Exponential; the consecutive outputs have a constant ratio of $2$.

Linear vs exponential comparison 3

Classify each function as linear or exponential. a) $f(x)=4x+7$ b) $g(x)=4(7)^x$

Hints

- Look at where the variable appears in each formula. - Compare the standard forms of a linear function and an exponential function.

Solution

1. In $f(x)=4x+7$, the variable is multiplied by a constant, so the function is linear. 2. In $g(x)=4(7)^x$, the variable is in the exponent, so the function is exponential.

Answer

a) Linear b) Exponential

Linear vs exponential comparison 4

A quantity starts at $50$. Situation A adds $5$ each week. Situation B doubles each week. Which situation is linear, and which is exponential?

Hints

- Ask whether the weekly change is additive or multiplicative. - “Doubles” describes multiplication by a constant factor.

Solution

1. Situation A uses the same additive change each week, so it is linear. 2. Situation B multiplies by the same factor $2$ each week, so it is exponential.

Answer

Situation A is linear. Situation B is exponential.

Linear vs exponential comparison 5

Which pattern is linear and which is exponential? Explain how the outputs change. <table> <tr><th>Step</th><td>$0$</td><td>$1$</td><td>$2$</td><td>$3$</td></tr> <tr><th>A</th><td>$8$</td><td>$12$</td><td>$16$</td><td>$20$</td></tr> <tr><th>B</th><td>$8$</td><td>$12$</td><td>$18$</td><td>$27$</td></tr> </table>

Hints

- The first two columns alone do not distinguish the rules. - Compare both the differences and the ratios across all the steps.

Solution

1. For A, each output increases by $4$; this is a constant difference, so A is linear. 2. For B, each output is $1.5$ times the previous one; this is a constant ratio, so B is exponential. 3. They agree for the first two steps but follow different rules afterward.

Answer

A is linear (add $4$); B is exponential (multiply by $1.5$).

Linear vs exponential comparison 6

<table> <tr><th>$x$</th><td>$0$</td><td>$2$</td><td>$4$</td><td>$6$</td></tr> <tr><th>$y$</th><td>$10$</td><td>$20$</td><td>$40$</td><td>$80$</td></tr> </table> Is the pattern linear or exponential? Explain using the changes in output.

Hints

- Compare outputs over equal intervals in $x$. - What would the differences look like if the pattern were linear?

Solution

1. The output differences for equal input increases are $10$, $20$, and $40$, so they are not constant. 2. The ratios are all $2$, so the output doubles whenever $x$ increases by $2$. 3. Therefore, the relationship is exponential, not linear.

Answer

Exponential. The output doubles for every $2$-unit increase in $x$; the differences do not stay constant.

Linear vs exponential comparison 7

At the beginning of an observation period, an algae culture covers $400\,\text{cm}^2$. After $2$ days, it covers $600\,\text{cm}^2$, and after $4$ days, it covers $900\,\text{cm}^2$. a) Use calculations to decide whether a linear or exponential growth model better fits the data. b) Write an equation for the appropriate model. c) Predict the area after a total of $1$ week if the growth pattern continues.

Hints

- Compare both the differences and the ratios between measurements over equal time intervals. - Use the initial value and the growth factor for each $2$-day interval. - Determine how many $2$-day growth intervals occur in $7$ days.

Solution

1. For a linear model, equal time intervals should have equal differences. Here, $600-400=200$, but $900-600=300$, so the data are not linear. 2. For an exponential model, equal time intervals should have equal ratios. Here, $\frac{600}{400}=1.5$ and $\frac{900}{600}=1.5$, so an exponential model fits. 3. The initial value is $400$, and the growth factor for every $2$ days is $1.5$. Thus, $f(t)=400(1.5)^{t/2}$, where $t$ is measured in days. Equivalently, $f(t)\approx 400(1.2247)^t$. 4. After $7$ days, $f(7)=400(1.5)^{7/2}\approx 1653.41$.

Answer

a) Exponential, because the ratios for equal time intervals are constant while the differences are not b) $f(t)=400(1.5)^{t/2}$, or approximately $f(t)=400(1.2247)^t$ c) About $1653.41\,\text{cm}^2$

Linear vs exponential comparison 8

Two cultures each start with $100$ organisms. Culture A grows by $50$ organisms per hour. Culture B doubles each hour. a) Write a linear function for A and an exponential function for B. b) Compare their populations after $2$ hours. c) At which whole-number hour do their populations first differ? Explain.

Hints

- Which culture changes by adding a fixed amount? - Which culture changes by repeated multiplication? - To find the first different hour, check the starting hour and then the next whole hour.

Solution

1. Adding $50$ organisms each hour gives $A(t)=100+50t$. 2. Doubling each hour gives $B(t)=100(2)^t$. 3. At $t=2$, $A(2)=200$ and $B(2)=400$. B has $200$ more organisms. 4. At $t=0$, both have $100$; at $t=1$, A has $150$ and B has $200$. They first differ at hour $1$.

Answer

a) $A(t)=100+50t$; $B(t)=100(2)^t$ b) $A(2)=200$, $B(2)=400$; B has $200$ more. c) Hour $1$, since A has $150$ and B has $200$ then.

Linear vs exponential comparison 9

Examine the two tables. For each table, decide whether the pattern is linear or exponential, and fill in the missing values. a) <table> <tr> <td>Time $t$</td> <td>$0$</td> <td>$1$</td> <td>$2$</td> <td>$3$</td> <td>$4$</td> </tr> <tr> <td>Amount $A$</td> <td>$200$</td> <td>$240$</td> <td></td> <td>$345.6$</td> <td></td> </tr> </table> b) <table> <tr> <td>Time $t$</td> <td>$0$</td> <td>$1$</td> <td>$2$</td> <td>$3$</td> <td>$4$</td> </tr> <tr> <td>Amount $B$</td> <td>$200$</td> <td>$240$</td> <td></td> <td></td> <td>$360$</td> </tr> </table>

Hints

- Test whether consecutive differences are constant. - Test whether consecutive ratios are constant. - Decide whether addition or multiplication takes one value to the next. - Use the known value at the end of each table to check the pattern.

Solution

1. In table a, $\frac{240}{200}=1.2$. Also, $240\cdot(1.2)^2=345.6$, so the constant growth factor is $1.2$, and the pattern is exponential. 2. The missing values in table a are $240\cdot1.2=288$ at $t=2$ and $345.6\cdot1.2=414.72$ at $t=4$. 3. In table b, $240-200=40$, and $200+4\cdot40=360$. The constant difference is $40$, so the pattern is linear. 4. The missing values in table b are $240+40=280$ at $t=2$ and $280+40=320$ at $t=3$.

Answer

a) Exponential; $288$ at $t=2$ and $414.72$ at $t=4$ b) Linear; $280$ at $t=2$ and $320$ at $t=3$

Linear vs exponential comparison 10

Compare two savings models. Both begin with $\$500$ at $t=0$. Model 1 increases by $\$25$ each year. Model 2 increases by $4\%$ each year. Identify the type of growth in each model, and find the balance after $5$ years.

Hints

- Distinguish between increasing by a fixed amount and increasing by a fixed percent. - A constant absolute change produces what type of model? - A change based on the current value produces what type of model? - Convert the percent increase to a growth factor.

Solution

1. Model 1 has a constant dollar increase, so it is linear. Its balance after $5$ years is $500+5\cdot25=625$. 2. Model 2 has a constant percent increase, so it is exponential with growth factor $1.04$. Its balance after $5$ years is $500\cdot(1.04)^5\approx608.33$.

Answer

Model 1: linear growth; $\$625.00$ after $5$ years Model 2: exponential growth; $\$608.33$ after $5$ years

Linear vs exponential comparison 11

The table shows weekly views of a new video. <table><tr><th>Week</th><td>$0$</td><td>$1$</td><td>$2$</td><td>$3$</td></tr><tr><th>Views</th><td>$200$</td><td>$300$</td><td>$450$</td><td>$675$</td></tr></table> a) Show that the views follow an exponential pattern, and give the weekly factor. b) Suppose instead that the number of views increased by the same number each week. Use weeks $0$ and $1$ to find that weekly increase. c) What would the linear pattern predict for week $3$? How many fewer views is that than the table shows?

Hints

- Compare neighboring entries using both division and subtraction. - For the fixed-number model, use the change over the first week. - Extend that same fixed change to week $3$, then compare predictions.

Solution

1. Consecutive ratios are $\frac{300}{200}=\frac{450}{300}=\frac{675}{450}=1.5$, so the pattern is exponential with factor $1.5$. 2. The change from week $0$ to week $1$ is $300-200=100$. A linear model based on those weeks adds $100$ weekly. 3. The linear prediction for week $3$ is $200+3\cdot100=500$ views, which is $675-500=175$ fewer than the actual count.

Answer

a) Exponential; weekly factor $1.5$ b) An increase of $100$ views each week c) $500$ views, which is $175$ fewer than $675$

Linear vs exponential comparison 12

The graph shows two lines and two exponential curves labeled $p$, $q$, $r$, and $s$, all passing through the point $P(0, 1)$. Match each equation to its graph. Use the characteristics of linear and exponential growth. (a) $y=1.5x+1$ (b) $y=0.5x+1$ (c) $y=(1.2)^x$ (d) $y=2^x$

Hints

- How can you distinguish a line from an exponential curve? - Compare the slopes of the two lines. - Compare how quickly the two curves become steeper. - Consider which functions grow fastest for large positive $x$.

Solution

1. Graphs $p$ and $r$ are lines, so they represent the linear functions. Graphs $q$ and $s$ are curved, so they represent the exponential functions. 2. Graph $p$ is steeper than graph $r$. Since $1.5>0.5$, graph $p$ matches (a), and graph $r$ matches (b). 3. Graph $q$ grows faster than graph $s$. Since $2>1.2$, graph $q$ matches (d), and graph $s$ matches (c).

Answer

$p$ $\rightarrow$ (a); $q$ $\rightarrow$ (d); $r$ $\rightarrow$ (b); $s$ $\rightarrow$ (c)

Linear vs exponential comparison 13

Let $f(x)=2^x$. The three panels show linear graphs together with $f$. Which linear graph intersects $f$ at more than one point?

Hints

- Look for places where a line switches from above the curve to below it. - A crossing may fall between numbered gridlines, so check between whole-number inputs too.

Solution

1. In panel a), the line $g_a(x)=x+2$ meets the curve at $(2, 4)$ and crosses it again between $x=-2$ and $x=0$, as seen on the graph. 2. In panel b), the line $g_b(x)=-x+1$ crosses $f$ only at $(0, 1)$. 3. In panel c), the line $g_c(x)=0.5x$ stays below $f$ on the displayed graph. 4. Therefore panel a) has more than one intersection.

Answer

Panel a)

Linear vs exponential comparison 14

The graph shows two models, $f$ in blue and $g$ in red. One model is linear, and the other is exponential. a) Which graph represents each type of growth? b) For what interval with $x>0$ does the linear model have greater values than the exponential model?

Hints

- Which graph is a line, and which graph is curved? - Locate the points where the graphs intersect. - Between the intersection points, compare which graph lies higher.

Solution

1. The red graph $g$ is a line, so it represents linear growth. The blue graph $f$ is curved and becomes steeper, so it represents exponential growth. 2. The functions intersect at $x=0$, because $f(0)=3^0=1$ and $g(0)=2\cdot0+1=1$. They also intersect at $x=1$, because $f(1)=3$ and $g(1)=3$. 3. Between the intersection points, the line is above the exponential curve. For example, at $x=0.5$, $g(0.5)=2$, while $f(0.5)=\sqrt{3}\approx1.732$. 4. Therefore, the linear model has greater values for $0<x<1$.

Answer

a) The red graph $g$ is linear, and the blue graph $f$ is exponential. b) $(0, 1)$

Linear vs exponential comparison 15

Two online communities each begin with $100$ members. Community A gains $12$ members each week. Community B increases its current membership by $12\%$ each week. Write a function for each community after $t$ weeks, and identify which model is linear and which is exponential.

Hints

- Separate a fixed numerical increase from a fixed percent increase. - Which situation uses repeated addition of the same amount? - Which situation multiplies the current amount by the same factor each week?

Solution

1. Community A adds the constant amount $12$ each week, so $A(t)=100+12t$. This is linear. 2. Community B retains the current amount and adds $12\%$ of it, giving growth factor $1.12$. Thus $B(t)=100(1.12)^t$. This is exponential.

Answer

Community A: $A(t)=100+12t$, linear Community B: $B(t)=100(1.12)^t$, exponential

Linear vs exponential comparison 16

a) Read the outputs of curve $s$ at $x=0$, $1$, and $2$. What factor relates each pair of neighboring outputs? b) Which graph is linear? Use the same horizontal steps to explain your choice.

Hints

- Compare points directly above neighboring whole-number inputs. - Is the change across a one-unit step best described by addition or multiplication? - Test the other graph over the same intervals.

Solution

1. Curve $s$ passes through $(0,2)$, $(1,4)$, and $(2,8)$, so its outputs double each time $x$ increases by $1$. 2. Curve $r$ adds $3$ each step, from $2$ to $5$ to $8$; it is linear. Curve $s$ is exponential because it has a fixed factor.

Answer

a) $s$: outputs $2,4,8$; factor $2$. b) $r$ is linear; it adds $3$ for each unit increase in $x$.

Linear vs exponential comparison 17

Two online puzzles start with the same number of plays. One gains $3$ plays each hour; the other doubles its total plays each hour. Match each rule to a curve. Explain your answer using the numbers of plays at hours $1$ and $2$.

Hints

- Read both heights at two consecutive whole-number hours. - Is each graph changing by the same amount or the same factor? - The curves can share an output without following the same rule.

Solution

1. Curve $u$ rises from $6$ at hour $1$ to $9$ at hour $2$, an increase of $3$; it matches the fixed addition. 2. Curve $v$ rises from $6$ to $12$ over the same interval; it matches doubling. 3. Both models show $6$ at hour $1$, but their later outputs separate.

Answer

Curve $u$: gain $3$ plays per hour. Curve $v$: double each hour. From hour $1$ to $2$, $u$ goes from $6$ to $9$ plays, while $v$ goes from $6$ to $12$ plays.

Linear vs exponential comparison 18

Nora compares $L(t)=10+10t$ and $E(t)=10(2)^t$. They have the same values at $t=0$ and $t=1$, so she says the two functions must be identical. Is her conclusion correct? Use another input to support your answer.

Hints

- A claim about every input can be disproved by one counterexample. - Choose an input other than the two already compared.

Solution

1. The models both give $10$ at $t=0$ and $20$ at $t=1$. 2. At $t=2$, $L(2)=10+20=30$, while $E(2)=10\cdot4=40$. 3. The functions are not identical; matching at two inputs does not make their rules the same.

Answer

No. At $t=2$, $L(2)=30$ and $E(2)=40$.

Linear vs exponential comparison 19

Jordan says, “A formula with the variable in the exponent must increase as the input increases.” Consider $E(t)=4(0.5)^t$ and $L(t)=4+2t$. a) Find both functions' values at $t=0$, $1$, and $2$. b) Describe how each changes from one step to the next. Is Jordan right?

Hints

- A small table can help you track how both functions change as the input increases. - Compare neighboring outputs by subtraction and by division. Which kind of change stays the same for each function?

Solution

1. The exponential values are $E(0)=4$, $E(1)=2$, and $E(2)=1$. 2. The linear values are $L(0)=4$, $L(1)=6$, and $L(2)=8$. 3. The exponential output is halved at each step, while the linear output increases by $2$. Jordan's claim is false: an exponential function with a base between $0$ and $1$ decreases.

Answer

a) $E(0)=4$, $E(1)=2$, $E(2)=1$; $L(0)=4$, $L(1)=6$, $L(2)=8$ b) $E$ is halved at each step; $L$ increases by $2$ each step. Jordan is wrong because an exponential function can decrease.

Linear vs exponential comparison 20

A bean plant is measured each week, to the nearest centimeter. It is $20\,\text{cm}$ tall at week $0$, $24\,\text{cm}$ at week $1$, and $35\,\text{cm}$ at week $3$. These models agree with the first two measurements: $L(t)=20+4t$ and $E(t)=20(1.2)^t$. Which model also matches the third measurement when rounded to the nearest centimeter? Check both predictions.

Hints

- In the exponential rule, another week means another multiplication by the same factor. How can you use that to work toward week $3$? - A height reported as $35\,\text{cm}$ may not be exactly $35\,\text{cm}$. Which predictions would round to that measurement?

Solution

1. At week $3$, the linear model predicts $L(3)=20+4(3)=32$. 2. The exponential model predicts $E(3)=20(1.2)^3=34.56$. 3. Rounded to the nearest centimeter, the predictions are $32\,\text{cm}$ and $35\,\text{cm}$. The exponential model matches the third measurement.

Answer

$E(t)=20(1.2)^t$. At week $3$, $E(3)=34.56\,\text{cm}\approx35\,\text{cm}$, while $L(3)=32\,\text{cm}$.

Linear vs exponential comparison 21

Two school videos are posted at the same time. Video A starts with $50$ views and gains $10$ views each hour. Video B starts with $40$ views and grows by $50\%$ each hour. Write a model for each video's views at whole-number hours $t$. At which first hour does B have more views than A? Check the earlier hours.

Hints

- Begin with the starting hour because the question asks when B first has more views. - A tie is not the same as B having more views. - Compare the models at successive whole-number hours.

Solution

1. The models are $A(t)=50+10t$ and $B(t)=40(1.5)^t$. 2. At hour $0$, A has $50$ views and B has $40$; B has fewer. 3. At hour $1$, both have $60$ views. 4. At hour $2$, A has $70$ views and B has $90$; B first has more views at hour $2$.

Answer

$A(t)=50+10t$; $B(t)=40(1.5)^t$. First overtaking: hour $2$. At hours $0$ and $1$, B has fewer or equal views.

Linear vs exponential comparison 22

The value of a collectible car increases from $\$20{,}000$ to $\$50{,}000$ over $10$ years. a) Find its value after $5$ years if it grows by a constant annual percent. b) Find its value after $5$ years if it increases by a constant dollar amount each year. c) Which model gives the greater value at $5$ years? Explain how the two models use the starting and current values differently. Round money to the nearest cent.

Hints

- How is a constant dollar change different from a constant percentage change? - Use the total dollar increase to find the linear model's halfway value. - Think about how a constant annual multiplier connects the starting value to the ten-year value.

Solution

1. In the percent-growth model, the ten-year growth factor is $\frac{50000}{20000}=2.5$, so the five-year factor is $\sqrt{2.5}$. The halfway value is $20000\sqrt{2.5}\approx\$31{,}622.78$. 2. In the linear model, the value rises by $\frac{50000-20000}{10}=3000$ dollars per year. After five years, it is $20000+5\cdot3000=\$35{,}000$. 3. The linear model gives the higher halfway value. It adds the same dollar amount each year. The exponential model grows by a fixed percent of the current value each year.

Answer

a) $\$31{,}622.78$ b) $\$35{,}000.00$ c) The linear model is higher. It adds the same amount each year; the percent model increases by a fixed share of the current value.

Linear vs exponential comparison 23

An experiment records $x_0=100$, $x_1=150$, and $x_2=225$ at times $t=0$, $t=1$, and $t=2$ hours. a) Show numerically that the data follow an exponential pattern. Why do they not follow a linear pattern? b) What would $x_2$ be if the change were linear while $x_0$ and $x_1$ stayed the same? c) Find the predicted value after $5$ hours for both the linear and exponential models.

Hints

- Compare consecutive differences and consecutive ratios. - A linear pattern has a constant difference; an exponential pattern has a constant ratio. - Use $x_n=x_0+nd$ for the linear model and $x_n=x_0r^n$ for the exponential model.

Solution

1. The consecutive ratios are $\frac{x_1}{x_0}=\frac{150}{100}=1.5$ and $\frac{x_2}{x_1}=\frac{225}{150}=1.5$. The constant ratio shows exponential growth. The consecutive differences are $150-100=50$ and $225-150=75$, so the pattern is not linear. 2. A linear model based on $x_0$ and $x_1$ has common difference $50$. Therefore, its next value would be $x_2=150+50=200$. 3. For the linear model, $x_5=100+5\cdot50=350$. For the exponential model, $x_5=100\cdot(1.5)^5=759.375$.

Answer

a) The ratios are both $1.5$, while the differences are $50$ and $75$. Therefore, the pattern is exponential, not linear. b) $200$ c) Linear: $350$; exponential: $759.375$

Linear vs exponential comparison 24

Two investment accounts begin with the same balance of $\$800$. Account $K_1$ grows linearly, and account $K_2$ grows exponentially. After $2$ years, both accounts have a balance of $\$1800$. a) Write a function for each account's balance after $t$ years. b) Determine which account has the greater balance after $1$ year, and find the difference between the balances.

Hints

- Start with the general forms of a linear model and an exponential model. - Use the value after $2$ years to determine each model's unknown rate. - Evaluate both functions at $t=1$.

Solution

1. For the linear account, write $K_1(t)=800+dt$. Since $K_1(2)=1800$, $800+2d=1800$, so $d=500$. Thus $K_1(t)=800+500t$. 2. For the exponential account, write $K_2(t)=800b^t$. Since $K_2(2)=1800$, $800b^2=1800$, so $b^2=2.25$ and $b=1.5$. Thus $K_2(t)=800(1.5)^t$. 3. After $1$ year, $K_1(1)=1300$ and $K_2(1)=1200$. The linear account is greater by $1300-1200=100$.

Answer

a) $K_1(t)=800+500t$; $K_2(t)=800(1.5)^t$ b) Account $K_1$ has the greater balance after $1$ year, by $\$100$.

Linear vs exponential comparison 25

Compare two savings plans. Plan A starts with $\$2000$ and increases by $\$150$ each year. Plan B starts with $\$1500$ and earns $5\%$ interest each year, compounded annually. a) Write a function for the balance in each plan after $t$ years. Identify each model as linear or exponential. b) Find the balance in each plan after $10$ years. c) By testing whole-number values of $t$ or making a table, determine the first full year when Plan B has a greater balance than Plan A.

Hints

- Decide which plan adds a fixed amount and which multiplies by a fixed growth factor. - Convert a $5\%$ annual increase to a growth factor. - Continue a table of whole-year balances until the inequality reverses.

Solution

1. Plan A has a constant yearly increase, so $A(t)=2000+150t$ is linear. Plan B has a yearly growth factor of $1.05$, so $B(t)=1500(1.05)^t$ is exponential. 2. After $10$ years, $A(10)=2000+150\cdot10=3500$, and $B(10)=1500\cdot(1.05)^{10}\approx2443.34$. 3. Testing whole-number years from $t=0$ upward shows that Plan A is at least as large through $t=30$. At $t=30$, $A(30)=6500$ and $B(30)\approx6482.91$, so Plan A is still greater. At $t=31$, $A(31)=6650$ and $B(31)\approx6807.06$, so Plan B is greater for the first time.

Answer

a) $A(t)=2000+150t$, linear; $B(t)=1500(1.05)^t$, exponential b) Plan A: $\$3500.00$; Plan B: $\$2443.34$ c) After $31$ years

Linear vs exponential comparison 26

Two plant cultures are grown under different conditions. Culture 1 initially covers $500\,\text{cm}^2$, and its area increases by $200\,\text{cm}^2$ each week. Culture 2 initially covers $50\,\text{cm}^2$, and its area increases by $50\%$ each week. a) Complete the table. Round Culture 2's areas to the nearest whole square centimeter. <table> <tr> <th>Week $t$</th> <th>$0$</th> <th>$1$</th> <th>$2$</th> <th>$3$</th> <th>$4$</th> <th>$5$</th> </tr> <tr> <th>Culture 1 in $\text{cm}^2$</th> <td>$500$</td> <td></td> <td></td> <td></td> <td></td> <td></td> </tr> <tr> <th>Culture 2 in $\text{cm}^2$</th> <td>$50$</td> <td></td> <td></td> <td></td> <td></td> <td></td> </tr> </table> b) Use the changes in area to explain which culture grows linearly and which grows exponentially. c) After how many full weeks does Culture 2 first cover a greater area than Culture 1?

Hints

- Determine whether each week's change is a fixed amount or a fixed percent. - Write a rule for each row before filling the table. - Continue the models beyond week $5$ until Culture 2 becomes greater.

Solution

1. Culture 1 follows $A(t)=500+200t$, giving $500, 700, 900, 1100, 1300, 1500$. 2. Culture 2 follows $B(t)=50(1.5)^t$. Its rounded values are $50, 75, 113, 169, 253, 380$. 3. Culture 1 is linear because its consecutive differences are always $200$. Culture 2 is exponential because each exact value is $1.5$ times the preceding value. 4. Continuing whole-number weeks beyond week $5$ shows that Culture 1 remains greater through week $9$. At week $9$, $A(9)=2300$ and $B(9)\approx1922.17$. At week $10$, $A(10)=2500$ and $B(10)\approx2883.25$. Therefore, Culture 2 first exceeds Culture 1 after $10$ full weeks.

Answer

a) Culture 1: $500, 700, 900, 1100, 1300, 1500$ Culture 2: $50, 75, 113, 169, 253, 380$ b) Culture 1 is linear because it has a constant difference. Culture 2 is exponential because it has a constant growth factor. c) $10$ weeks

Linear vs exponential comparison 27

You can choose between two pay plans for a $14$-day summer job. Plan 1: You earn $\$30.00$ on the first day. Each following day's pay is $\$15.00$ more than the previous day's pay. Plan 2: You earn only $\$0.02$ on the first day. Each following day's pay is twice the previous day's pay. a) Find the amount you would earn on day $14$ under each plan. Which plan pays more on that day? b) Write a function $f(n)$ for the daily pay under each plan, where $n$ is the day number. c) Identify the linear model and the exponential model. Briefly justify your choices.

Hints

- Decide whether the pay changes by adding the same amount or multiplying by the same factor. - List the first three days under each plan to identify the pattern. - Use an exponent to represent repeated multiplication. - Remember that no increase or doubling has occurred yet on day $1$.

Solution

1. Plan 1 has a constant daily increase of $15$, so $f_1(n)=30+15(n-1)=15n+15$. On day $14$, $f_1(14)=225$. 2. Plan 2 has a constant growth factor of $2$, so $f_2(n)=0.02(2)^{n-1}$. On day $14$, $f_2(14)=0.02\cdot2^{13}=163.84$. 3. Since $225>163.84$, Plan 1 pays more on day $14$. Plan 1 is linear because it has a constant difference, and Plan 2 is exponential because it has a constant ratio.

Answer

a) Plan 1: $\$225.00$; Plan 2: $\$163.84$. Plan 1 pays more on day $14$. b) $f_1(n)=15n+15$; $f_2(n)=0.02(2)^{n-1}$ c) Plan 1 is linear because the daily difference is constant. Plan 2 is exponential because the daily growth factor is constant.

Linear vs exponential comparison 28

Two newly formed clubs have different membership patterns. Club A starts with $200$ members and gains $40$ members each month. Club B starts with $50$ members, and its membership increases by $15\%$ each month. a) Write a function $M(t)$ for each club's membership $t$ months after the clubs are formed. b) Find each club's membership after $12$ months and after $24$ months. Round Club B's membership to the nearest whole number. c) Explain which club has linear growth and which has exponential growth.

Hints

- Convert the $15\%$ increase to a growth factor. - Distinguish between a fixed number of new members and a fixed percent increase. - Substitute each time value into the appropriate function.

Solution

1. Club A gains a constant number of members, so $M_A(t)=200+40t$. 2. Club B has a monthly growth factor of $1.15$, so $M_B(t)=50(1.15)^t$. 3. After $12$ months, $M_A(12)=680$, and $M_B(12)=50\cdot(1.15)^{12}\approx267.51$, or about $268$ members. 4. After $24$ months, $M_A(24)=1160$, and $M_B(24)=50\cdot(1.15)^{24}\approx1431.26$, or about $1431$ members. 5. Club A is linear because its absolute monthly change is constant. Club B is exponential because its monthly percent change is constant.

Answer

a) $M_A(t)=200+40t$; $M_B(t)=50(1.15)^t$ b) After $12$ months: Club A, $680$; Club B, about $268$. After $24$ months: Club A, $1160$; Club B, about $1431$. c) Club A has linear growth, and Club B has exponential growth.

Linear vs exponential comparison 29

An initial investment of $\$10{,}000$ is held for several years. Compare the two plans. Plan A: A fixed $\$320$ is added at the end of each year. The added amounts do not earn additional interest. Plan B: The account earns $3\%$ interest per year, compounded annually. a) Find the balance in each plan after $1$, $2$, and $5$ years. b) Identify the linear model and the exponential model. Justify your answer using the yearly increases. c) Compare the balances after $20$ years. Which plan gives the greater balance?

Hints

- Decide whether each plan adds the same dollar amount or applies a percent to the current balance. - Write a direct formula for the balance after $t$ years. - Compare the yearly differences in the two models. - Consider what happens when the amount earning interest becomes larger each year.

Solution

1. Plan A follows $A(t)=10{,}000+320t$. Its balances are $A(1)=10{,}320$, $A(2)=10{,}640$, and $A(5)=11{,}600$. 2. Plan B follows $B(t)=10{,}000(1.03)^t$. Its balances are $B(1)=10{,}300$, $B(2)=10{,}609$, and $B(5)\approx11{,}592.74$. 3. Plan A is linear because the yearly increase is always $\$320$. Plan B is exponential because the balance is multiplied by $1.03$ each year, so its dollar increases become larger. 4. After $20$ years, $A(20)=16{,}400$, and $B(20)=10{,}000\cdot(1.03)^{20}\approx18{,}061.11$. Plan B gives the greater balance.

Answer

a) Plan A: $\$10{,}320.00$, $\$10{,}640.00$, and $\$11{,}600.00$. Plan B: $\$10{,}300.00$, $\$10{,}609.00$, and $\$11{,}592.74$. b) Plan A is linear; Plan B is exponential. c) After $20$ years, Plan A has $\$16{,}400.00$, and Plan B has about $\$18{,}061.11$. Plan B gives the greater balance.

Linear vs exponential comparison 30

Two biological cultures are observed under different conditions. At $t=0$, each culture contains $800$ organisms. Culture 1 increases each hour by $15\%$ of its population from the previous hour. Culture 2 gains $150$ organisms each hour. a) Write a function $N(t)$ for each culture's population after $t$ hours. b) Find both populations after $3$ hours and after $8$ hours. Round nonwhole populations to the nearest whole organism. c) Explain how the hourly increases in the two cultures change over time. Why will Culture 1 eventually exceed Culture 2, even though it grows more slowly at first?

Hints

- Determine whether each culture adds a fixed amount or multiplies by a fixed factor. - Use the general forms $y=mx+b$ and $y=ab^x$. - Compare the absolute increase from one hour to the next in each model. - Use the values at $3$ and $8$ hours to observe how the comparison changes.

Solution

1. Culture 1 has exponential growth, so $N_1(t)=800(1.15)^t$. Culture 2 has linear growth, so $N_2(t)=800+150t$. 2. After $3$ hours, $N_1(3)=1216.7\approx1217$, and $N_2(3)=1250$. 3. After $8$ hours, $N_1(8)\approx2447.22\approx2447$, and $N_2(8)=2000$. 4. Culture 2 always gains $150$ organisms per hour. Culture 1 gains $15\%$ of an increasing population, so its absolute hourly increase becomes larger over time. A growing exponential model with factor greater than $1$ eventually exceeds a linear model.

Answer

a) $N_1(t)=800(1.15)^t$; $N_2(t)=800+150t$ b) After $3$ hours: Culture 1, about $1217$; Culture 2, $1250$. After $8$ hours: Culture 1, about $2447$; Culture 2, $2000$. c) Culture 2 has a constant hourly increase. Culture 1's absolute hourly increase grows because $15\%$ is taken from an increasingly large population.

Linear vs exponential comparison 31

An initial investment of $\$12{,}000$ is placed in two different accounts. Account 1 has balances of $\$13{,}440.00$ after $2$ years, $\$14{,}880.00$ after $4$ years, and $\$16{,}320.00$ after $6$ years. Account 2 has balances of $\$13{,}483.20$ after $2$ years, $\$15{,}149.72$ after $4$ years, and $\$17{,}022.23$ after $6$ years. a) Determine whether each account follows a linear or exponential model. b) Find the yearly dollar increase for the linear model and the annual interest rate for the exponential model. c) Find the balance in each account after $10$ years.

Hints

- Check whether equal time intervals produce equal differences or equal ratios. - Find the differences between consecutive balances. - Find the ratios between consecutive balances. - The data are given in $2$-year intervals, but part b asks for yearly growth.

Solution

1. For Account 1, the balance increases by $1440$ every $2$ years: $13{,}440-12{,}000=1440$, $14{,}880-13{,}440=1440$, and $16{,}320-14{,}880=1440$. Therefore, it is linear, with a yearly increase of $1440\div2=720$. 2. For Account 2, the consecutive $2$-year growth factors are $\frac{13{,}483.20}{12{,}000}=1.1236$, $\frac{15{,}149.72}{13{,}483.20}\approx1.1236$, and $\frac{17{,}022.23}{15{,}149.72}\approx1.1236$. Thus the model is exponential. The yearly growth factor satisfies $b^2=1.1236$, so $b=1.06$, giving an annual interest rate of $6\%$. 3. After $10$ years, Account 1 has $12{,}000+10\cdot720=19{,}200$. Account 2 has $12{,}000\cdot(1.06)^{10}\approx21{,}490.17$.

Answer

a) Account 1 is linear; Account 2 is exponential. b) Account 1 increases by $\$720.00$ per year. Account 2 earns $6\%$ per year. c) Account 1: $\$19{,}200.00$; Account 2: about $\$21{,}490.17$

Linear vs exponential comparison 32

Two candles, Candle A and Candle B, each begin with a height of $25\,\text{cm}$. Candle A burns at a constant rate, so its height decreases by $2\,\text{cm}$ each hour. Candle B's height decreases by $10\%$ of its current height each hour. a) Find the height of each candle after $5$ hours. Round any non-integer height to the nearest hundredth of a centimeter. b) Determine when Candle A reaches a height of $0\,\text{cm}$. c) Identify the linear decay model and the exponential decay model. Justify your answer using the rates of change.

Hints

- Decide whether each candle loses a fixed height or a fixed percent of its current height. - What factor represents a $10\%$ decrease? - Which model has a decreasing absolute loss over time? - Write a function for each candle.

Solution

1. Candle A follows $h_A(t)=25-2t$. After $5$ hours, $h_A(5)=25-2\cdot5=15\,\text{cm}$. 2. Candle B has decay factor $1-0.10=0.9$, so $h_B(t)=25(0.9)^t$. After $5$ hours, $h_B(5)=25\cdot(0.9)^5=14.76225\,\text{cm}\approx14.76\,\text{cm}$. 3. Candle A reaches $0\,\text{cm}$ when $25-2t=0$. Solving gives $t=12.5$ hours. 4. Candle A has linear decay because it loses a constant $2\,\text{cm}$ per hour. Candle B has exponential decay because it retains the constant factor $0.9$ each hour.

Answer

a) Candle A: $15\,\text{cm}$; Candle B: $14.76\,\text{cm}$ b) $12.5$ hours c) Candle A has linear decay because its hourly decrease is constant. Candle B has exponential decay because its hourly decay factor is constant.

Linear vs exponential comparison 33

Examine the three graphs labeled $p$, $q$, and $r$. One graph represents a linear function, and the other two represent exponential functions. a) Which graph is linear? Find its equation. b) Match the exponential graphs to the functions $f(x)=0.5(1.4)^x$ and $g(x)=2(0.7)^x$. c) Use an example from the graph to explain the difference between linear decay and exponential decay.

Hints

- What geometric shape is the graph of a linear function? - Compare the y-intercepts with the initial values in the exponential functions. - For part c, compare the displayed y-values at $x=0$ and $x=1$ on the two decreasing graphs.

Solution

1. Graph $q$ is a line. It has y-intercept $3$ and x-intercept $6$, so its slope is $\frac{0-3}{6-0}=-0.5$. Its equation is $y=-0.5x+3$. 2. Graph $p$ shows exponential decay and has y-intercept $2$, so it matches $g(x)=2(0.7)^x$. Graph $r$ shows exponential growth and has y-intercept $0.5$, so it matches $f(x)=0.5(1.4)^x$. 3. From $x=0$ to $x=1$, graph $q$ decreases from $3$ to $2.5$, a difference of $-0.5$. Over the same interval, graph $p$ decreases from $2$ to $1.4$, so its output is multiplied by $0.7$. This illustrates constant difference for linear decay and constant factor for exponential decay.

Answer

a) Graph $q$ is linear, with equation $y=-0.5x+3$. b) Graph $p$: $g(x)=2(0.7)^x$; Graph $r$: $f(x)=0.5(1.4)^x$ c) For example, from $x=0$ to $x=1$, $q$ changes from $3$ to $2.5$, a difference of $-0.5$, while $p$ changes from $2$ to $1.4$, a factor of $0.7$. Linear decay has a constant difference; exponential decay has a constant factor.

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Linear vs exponential comparison

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