Question 1: The value of a machine is modeled by the function $V(t) = 35000(0.85)^t$, where $t$ is the number of years since purchase. Which of the following best describes what the number $0.85$ represents in this context?

• A) The machine’s value decreases by $85\%$ each year.
• B) The machine’s value decreases by $15\%$ each year.
• C) The machine is worth $\$85$ after one year.
• D) The machine’s value increases by $15\%$ each year.

Question 2: A bacterial culture doubles in population every $3$ hours. If the initial population is $200$, write an expression representing the population after $h$ hours.

Question 3: The population of a town increases by $6\%$ every year. If the population in $2020$ was $15,000$, which expression gives the estimated population $t$ years after $2020$?

• A) $15000(0.06)^t$
• B) $15000(1.6)^t$
• C) $15000(1.06)^t$
• D) $15000(t)^{1.06}$

Question 1: The value of a machine is modeled by the function $V(t) = 35000(0.85)^t$, where $t$ is the number of years since purchase. Which of the following best describes what the number $0.85$ represents in this context?

In an exponential decay model $V(t) = P(1 – r)^t$, the base $0.85$ represents the retention factor ($1 – r$). Since $1 – 0.85 = 0.15$, it indicates a decrease of $15\%$ each year.

Answer: B) The machine’s value decreases by $15\%$ each year.

Question 2: A bacterial culture doubles in population every $3$ hours. If the initial population is $200$, write an expression representing the population after $h$ hours.

For doubling growth, the base is $2$, and the exponent is the number of hours $h$ divided by the doubling period ($3$ hours). Multiplying by the initial population $200$ gives:

Question 3: The population of a town increases by $6\%$ every year. If the population in $2020$ was $15,000$, which expression gives the estimated population…

Using the exponential growth formula $P = P_0(1 + r)^t$ with initial population $15,000$ and growth rate $0.06$, the model for $t$ years after $2020$ is $15000(1.06)^t$.

Question 1: The function $f(t) = 4500(1.08)^t$ gives the estimated value of an investment after $t$ years. Which of the following equations best displays the estimated value of the investment on a monthly basis?

• A) $f(t) = 4500(1.0064)^{12t}$
• B) $f(t) = 4500(1.08)^{\frac{t}{12}}$
• C) $f(t) = 4500(1.0006)^{12t}$
• D) $f(t) = 4500(1.08)^{12t}$

Question 2: A population of cells triples every $5$ hours. If the population is initially $100$, which expression represents the population after $h$ hours?

• A) $100(3)^{\frac{h}{5}}$
• B) $100(3)^{5h}$
• C) $300^{\frac{h}{5}}$
• D) $100(5)^{\frac{h}{3}}$

Question 3: The value of a car depreciates according to $V(t) = 25000(0.80)^t$. Which function shows the value of the car evaluated every month $m$, where $m = 12t$?

• A) $V(m) = 25000(0.80)^{\frac{m}{12}}$
• B) $V(m) = 25000(0.982)^m$
• C) $V(m) = 25000(0.067)^m$
• D) $V(m) = 25000(0.80)^{12m}$

Question 1: The function $f(t) = 4500(1.08)^t$ gives the estimated value of an investment after $t$ years. Which of the following equations best displays the estimated value of the investment on a monthly basis?

To convert the annual time scale ($t$ years) to a monthly one, the exponent must reflect $12$ months per year ($12t$). To keep the overall value equivalent, raise the annual growth factor to the power of $\frac{1}{12}$:

Answer: A) $f(t) = 4500(1.0064)^{12t}$

Question 2: A population of cells triples every $5$ hours. If the population is initially $100$, which expression represents the population after $h$ hours?

For tripling growth, the base is $3$, and the exponent is the number of hours $h$ divided by the period of tripling ($5$ hours). Multiplying by the initial population $100$ gives:

Answer: A) $100(3)^{\frac{h}{5}}$

Question 3: The value of a car depreciates according to $V(t) = 25000(0.80)^t$. Which function shows the value of the car evaluated every month $m$, where $m = 12t$?

Since $m = 12t$, we can solve for $t$ in terms of $m$: $t = \frac{m}{12}$.

Substitute $t = \frac{m}{12}$ into the original depreciation function $V(t)$:

This matches option A.

Answer: A) $V(m) = 25000(0.80)^{\frac{m}{12}}$

Question 1: Which of the following scenarios is best modeled by a linear function?

• A) The value of a car decreases by $12\%$ each year.
• B) A savings account earns $4\%$ interest compounded annually.
• C) A rental car company charges a flat fee of $\$50$ plus $\$0.20$ per mile driven.
• D) A population of bacteria triples every hour.

Question 2: Function $f$ is linear and has values $f(2) = 11$ and $f(4) = 19$. Function $g$ is exponential and has values $g(0) = 2$ and $g(1) = 6$. At what value of $x$ ($x > 0$) do their values first exceed each other?

Question 3: Which statement correctly compares linear growth and exponential growth for large positive values of $x$?

• A) Linear growth eventually exceeds exponential growth.
• B) Exponential growth eventually exceeds any linear growth model.
• C) Both growths increase at the exact same constant proportion.
• D) Linear growth accelerates faster than exponential growth.

Question 1: Which of the following scenarios is best modeled by a linear function?

A linear function models situations with a constant rate of change (constant addition/subtraction). Choice C features a fixed base fee plus a constant price per mile ($0.20$ per mile), which is linear. (Choices A, B, and D involve percentages or multiplicative factors, making them exponential).

Answer: C) A rental car company charges a flat fee of $\$50$ plus $\$0.20$ per mile driven.

Question 2: Function $f$ is linear and has values $f(2) = 11$ and $f(4) = 19$. Function $g$ is exponential and has values $g(0) = 2$ and $g(1) = 6$. At what value of $x$ ($x > 0$) do their values first exceed each other?

First, find the equation for linear function $f(x)$:

Using $f(2) = 11$, we get $4(2) + c = 11 \implies c = 3$, so $f(x) = 4x + 3$.

Next, find the equation for exponential function $g(x) = a(b)^x$:

Test integer values of $x > 0$:

• For $x = 1$: $f(1) = 7$, $g(1) = 6$ ($f > g$)
• For $x = 2$: $f(2) = 11$, $g(2) = 18$ ($g > f$, values cross between $1$ and $2$)
• For $x = 3$: $f(3) = 15$, $g(3) = 54$

Question 3: Which statement correctly compares linear growth and exponential growth…

The text continues below, but fundamentally, linear growth adds a fixed amount over equal intervals, whereas exponential growth multiplies by a constant factor, eventually growing much faster than any linear function.

Question 1: The mass of a radioactive sample is given by $m(t) = 800(0.5)^{\frac{t}{5}}$, where $t$ is the time in days. Which of the following represents the half-life of the sample, in days?

• A) $0.5$ days
• B) $5$ days
• C) $800$ days
• D) $400$ days

Question 2: A colony of bacteria triples in population every $4$ hours. If the initial population is $500$, which function models the population $p$ after $h$ hours?

• A) $p(h) = 500(3)^{\frac{h}{4}}$
• B) $p(h) = 500(3)^{4h}$
• C) $p(h) = 1500^{\frac{h}{4}}$
• D) $p(h) = 500\left(\frac{1}{3}\right)^{\frac{h}{4}}$

Question 3: A medication diminishes in the bloodstream with a half-life of $2$ hours. If a patient is given an initial dose of $400$ mg, how many milligrams remain in the bloodstream after $6$ hours?

Question 1: The mass of a radioactive sample is given by $m(t) = 800(0.5)^{\frac{t}{5}}$, where $t$ is the time in days. Which of the following represents the half-life of the sample, in days?

In the half-life decay formula $m(t) = m_0(0.5)^{\frac{t}{h}}$, the denominator in the exponent ($h$) is the half-life. Here, the exponent is $\frac{t}{5}$, meaning the half-life is $5$ days.

Answer: B) $5$ days

Question 2: A colony of bacteria triples in population every $4$ hours. If the initial population is $500$, which function models the population $p$ after $h$ hours?

For triple growth, the base is $3$, and the exponent is the time $h$ divided by the tripling period ($4$ hours). With initial population $500$:

Answer: A) $p(h) = 500(3)^{\frac{h}{4}}$

Question 3: A medication diminishes in the bloodstream with a half-life of $2$ hours…

The problem text continues below, but the general model for a half-life of $2$ hours with an initial dose $D_0$ is given by $D(t) = D_0(0.5)^{\frac{t}{2}}$.

Question 1: An account is opened with an initial deposit of $\$5000$ and earns $4\%$ interest compounded semi-annually. Which function gives the account balance $t$ years after the initial deposit?

• A) $f(t) = 5000(1.04)^{2t}$
• B) $f(t) = 5000(1.02)^{2t}$
• C) $f(t) = 5000(1.04)^{\frac{t}{2}}$
• D) $f(t) = 5000(1.08)^{2t}$

Question 2: The population of a city is modeled by $P(t) = 85000(1.025)^t$, where $t$ is the number of years since $2015$. What does the number $1.025$ signify in this context?

• A) The population increases by $25\%$ each year.
• B) The population increases by $2.5\%$ each year.
• C) The total population in year zero was $1,025$.
• D) The population multiplies by $1.025$ every decade.

Question 3: A digital asset loses $20\%$ of its user base every month. If it starts with $100,000$ users, write an expression for the user base after $m$ months.

Question 1: An account is opened with an initial deposit of $\$5000$ and earns $4\%$ interest compounded semi-annually. Which function gives the account balance $t$ years after the initial deposit?

Semi-annual compounding means interest is compounded $n = 2$ times per year. Using $A = P\left(1 + \frac{r}{n}\right)^{nt}$: $P = 5000$, $r = 0.04$, and $nt = 2t$. The rate per period is $\frac{0.04}{2} = 0.02$:

This matches choice B.

Answer: B) $f(t) = 5000(1.02)^{2t}$

Question 2: The population of a city is modeled by $P(t) = 85000(1.025)^t$, where $t$ is the number of years since 2015. What does the number $1.025$ signify in this context?

In an exponential growth model of the form $P(t) = P_0(b)^t$, the base $b$ represents the growth factor, equal to $1 + r$. Here, $1.025 = 1 + 0.025$, which means the population increases by $2.5\%$ each year.

Answer: B) The population increases by $2.5\%$ each year.

Question 1: The function $f(t) = 4500(1.035)^t$ gives the value of an investment after $t$ years. Which of the following best describes the meaning of the number $1.035$ in this context?

• A) The investment increases by $3.5\%$ each year.
• B) The investment increases by $35\%$ each year.
• C) The total value of the investment after one year is $\$3.50$.
• D) The annual interest rate is $1.035\%$.

Question 2: A rare coin’s value is modeled by $V(m) = 150(0.92)^m$, where $m$ is the number of months since January. Which statement is true based on the model?

• A) The initial value of the coin is $\$92$.
• B) The coin loses $92\%$ of its value each month.
• C) The coin retains $92\%$ of its value each month.
• D) The coin increases in value by $8\%$ each month.

Question 3: The population of a species of insect is given by $P(d) = 200(1.18)^d$, where $d$ is measured in days. By what percentage does the insect population increase every day?

Question 1: The function $f(t) = 4500(1.035)^t$ gives the value of an investment after $t$ years. Which of the following best describes the meaning of the number $1.035$ in this context?

In an exponential growth function $f(t) = a(1 + r)^t$, the base $1.035$ represents $1 + r$. The percentage growth rate is $0.035$, or $3.5\%$ increase each year.

Answer: A) The investment increases by $3.5\%$ each year.

Question 2: A rare coin’s value is modeled by $V(m) = 150(0.92)^m$, where $m$ is the number of months since January. Which statement is true based on the model?

The initial value when $m = 0$ is $150$. The base $0.92$ represents the retention factor ($1 – r$), meaning the coin retains $92\%$ of its value each month, which corresponds to losing $1 – 0.92 = 0.08$ ($8\%$) of its value each month.

Answer: C) The coin retains $92\%$ of its value each month. (Or evaluate remaining options accordingly depending on full list view).

Question 3: The population of a species of insect is given by $P(d) = 200(1.18)^d$, where $d$ is measured in days. By what percentage does the insect population increase every day?

In an exponential growth model $P(d) = a(1 + r)^d$, the base represents the growth factor $1 + r$. Here, the growth factor is $1.18$.

Set up the equation to find the growth rate $r$:

Convert the decimal rate $0.18$ to a percentage by multiplying by $100\%$, yielding an $18\%$ increase per day.

Answer: $18\%$

Question 1: Which of the following scenarios best describes an exponential growth model?

• A) A car loses $\$2,000$ in market value each year.
• B) A savings account earns $5\%$ interest compounded annually on its current balance.
• C) A worker earns $\$25$ for every hour worked.
• D) A train travels at a steady speed of $60$ miles per hour.

Question 2: The value of a collectible item increases by $12\%$ of its previous year’s value each year. Which type of equation best models the value of the item over time?

• A) Linear equation with a positive slope
• B) Linear equation with a negative slope
• C) Exponential growth function
• D) Quadratic function with a maximum value

Question 3: Explain why an exponential function will eventually exceed any linear function for sufficiently large positive values of $x$.

Question 1: Which of the following scenarios best describes an exponential growth model?

Exponential growth involves compounding by a constant percentage/factor over equal time intervals. Earning $5\%$ interest compounded annually means multiplying the current balance by $1.05$ each year.

Answer: B) A savings account earns $5\%$ interest compounded annually on its current balance.

Question 2: The value of a collectible item increases by $12\%$ of its previous year’s value each year. Which type of equation best models the value of the item over time?

Increasing by a percentage of the previous year’s value creates compound multiplication ($V(t) = V_0(1.12)^t$), which defines an exponential growth function.

Answer: C) Exponential growth function

Question 3: Explain why an exponential function will eventually exceed any linear function for sufficiently large positive values of $x$.

A linear function grows at a constant additive rate ($m$), while an exponential growth function ($b > 1$) grows at a multiplicative rate. As $x$ increases, multiplying by $b$ repeatedly accelerates growth continuously, ensuring the exponential function’s rate of increase eventually surpasses any constant additive slope $m$.

Question 1: If $4^{2x} = 8^{x+1}$, what is the value of $x$?

• A) $1$
• B) $2$
• C) $3$
• D) $4$

Question 2: The mass of a medication in a patient’s bloodstream decreases by half every $3$ hours. If the initial injection is $80$ milligrams, which function models the mass $M$, in milligrams, remaining after $t$ hours?

• A) $M(t) = 80\left(\frac{1}{2}\right)^{3t}$
• B) $M(t) = 80\left(\frac{1}{2}\right)^{\frac{t}{3}}$
• C) $M(t) = 40^{\frac{t}{3}}$
• D) $M(t) = 80(3)^{-\frac{t}{2}}$

Question 3: If $9^x = 27^{x-2}$, what is the value of $x$?

Question 1: If $4^{2x} = 8^{x+1}$, what is the value of $x$?

Express both bases with a common base of $2$ ($4 = 2^2$ and $8 = 2^3$):

Set the exponents equal and solve for $x$:

Answer: C) 3

Question 2: The mass of a medication in a patient’s bloodstream decreases by half every $3$ hours. If the initial injection is $80$ milligrams, which function models the mass $M$, in milligrams, remaining after $t$ hours?

For half-life decay, the base is $\frac{1}{2}$, and the exponent is the elapsed time $t$ divided by the half-life period ($3$ hours). Multiplying by the initial mass $80$ gives:

Answer: B) $M(t) = 80\left(\frac{1}{2}\right)^{\frac{t}{3}}$

Question 3: If $9^x = 27^{x-2}$, what is the value of $x$?

Rewrite both sides using a common base of $3$ ($9 = 3^2$ and $27 = 3^3$):

Equate the exponents and solve for $x$:

Question 1: An initial investment of $\$10,000$ is placed in an account that earns $4\%$ interest compounded semi-annually. Which of the following functions gives the account value $V$, in dollars, after $t$ years?

• A) $V(t) = 10000(1.04)^{2t}$
• B) $V(t) = 10000(1.02)^{2t}$
• C) $V(t) = 10000(1.04)^{\frac{t}{2}}$
• D) $V(t) = 10000(1.08)^{t}$

Question 2: A culture of bacteria triples in population every 6 hours. If the starting count is $450$ bacteria, which expression models the population $P$ after $h$ hours?

• A) $P(h) = 450(3)^{6h}$
• B) $P(h) = 450(3)^{\frac{h}{6}}$
• C) $P(h) = (450 \cdot 3)^{\frac{h}{6}}$
• D) $P(h) = 450\left(\frac{1}{3}\right)^{\frac{h}{6}}$

Question 3: Explain how to modify an annual growth function $y = a(b)^x$ so that it calculates monthly growth steps instead.

Question 1: An initial investment of $\$10,000$ is placed in an account that earns $4\%$ interest compounded semi-annually. Which of the following functions gives the account value $V$, in dollars, after $t$ years?

Semi-annual compounding means there are $2$ compounding periods per year ($2t$ total periods), and the rate per period is $\frac{0.04}{2} = 0.02$, giving a period multiplier of $1.02$:

Answer: B) $V(t) = 10000(1.02)^{2t}$

Question 2: A culture of bacteria triples in population every $6$ hours. If the starting count is $450$ bacteria, which expression models the population $P$ after $h$ hours?

The base is $3$ (tripling), and the exponent represents the number of $6$-hour cycles elapsed, which is $\frac{h}{6}$. Multiplying by the initial count $450$ gives:

Answer: B) $P(h) = 450(3)^{\frac{h}{6}}$

Question 3: Explain how to modify an annual growth function $y = a(b)^x$ so that it calculates monthly growth steps instead.

To convert an annual model ($x$ in years) to a monthly model, the exponent must account for $12$ months per year by multiplying $x$ by $12$ ($12x$). To maintain mathematical equivalence over the year, raise the annual growth factor $b$ to the power of $\frac{1}{12}$:

Question 1: Which of the following expressions is equivalent to $(2x^3)^4$?

• A) $2x^{12}$
• B) $8x^{12}$
• C) $16x^{7}$
• D) $16x^{12}$

Question 2: If $x > 0$, which expression is equivalent to $\sqrt{x^5} \cdot \sqrt[3]{x^2}$?

• A) $x^{\frac{19}{6}}$
• B) $x^{\frac{10}{6}}$
• C) $x^{\frac{7}{5}}$
• D) $x^{\frac{11}{6}}$

Question 3: Which expression is equivalent to $\frac{9^x}{3^{2x}}$ for all real values of $x$?

Question 1: Which of the following expressions is equivalent to $(2x^3)^4$?

Apply the power of a product rule by raising both the coefficient and the variable term to the fourth power: $(ab)^n = a^n b^n$:

Answer: D) $16x^{12}$

Question 2: If $x > 0$, which expression is equivalent to $\sqrt{x^5} \cdot \sqrt[3]{x^2}$?

Convert radical expressions into fractional exponent form ($\sqrt{x^5} = x^{\frac{5}{2}}$ and $\sqrt[3]{x^2} = x^{\frac{2}{3}}$):

Find a common denominator ($6$) to add the fractions ($\frac{15}{6} + \frac{4}{6} = \frac{19}{6}$):

Answer: A) $x^{\frac{19}{6}}$

Question 3: Which expression is equivalent to $\frac{9^x}{3^{2x}}$ for all real values of $x$?

Rewrite $9$ with a base of $3$ ($9 = 3^2$) so that the numerator becomes $(3^2)^x = 3^{2x}$:

Question 1: The population of a town is modeled by the function $P(t) = 12000(1.06)^t$, where $t$ is the number of years since 2020. Which of the following best describes the meaning of the number $1.06$ in this context?

• A) The population increases by $6$ people each year.
• B) The population increases by $6\%$ each year.
• C) The population is multiplied by $1.06$ every six years.
• D) The projected population in the year 2026 is $1.06$ times the 2020 population.

Question 2: A computer’s operating value decreases according to the equation $V = 1800(0.75)^t$, where $t$ is measured in years. Which of the following represents the annual rate of depreciation?

• A) $75\%$
• B) $25\%$
• C) $0.75\%$
• D) $1.75\%$

Question 3: Explain how to determine the initial amount and the percentage rate of growth from an exponential equation given in the form $y = a(b)^x$.

Question 1: The population of a town is modeled by the function $P(t) = 12000(1.06)^t$, where $t$ is the number of years since 2020. Which of the following best describes the meaning of the number $1.06$ in this context?

In an exponential growth model $P(t) = a(1 + r)^t$, the base $1.06$ represents $1 + r$. The growth rate $r = 1.06 – 1 = 0.06$, which means an increase of $6\%$ each year.

Answer: B) The population increases by $6\%$ each year.

Question 2: A computer’s operating value decreases according to the equation $V = 1800(0.75)^t$, where $t$ is measured in years. Which of the following represents the annual rate of depreciation?

The retention factor is $0.75$, meaning the item keeps $75\%$ of its value each year. The rate of depreciation (loss) is $1 – 0.75 = 0.25$, or $25\%$.

Answer: B) $25\%$

Question 3: Explain how to determine the initial amount and the percentage rate of growth from an exponential equation given in the form $y = a(b)^x$.

The initial amount is given directly by the constant coefficient $a$ (the value when $x = 0$). The percentage rate of growth is found by taking the growth factor $b$, subtracting $1$ ($r = b – 1$), and multiplying by $100\%$ to convert it to a percentage.