Exponential growth: Algebra 1 Regents Practice

On an island, a rabbit population doubles every month for two years. Which type of function best models this increase?

• (A) exponential growth ✓
• (B) linear growth
• (C) exponential decay
• (D) linear decay

Repeated multiplication by a fixed factor is exponential. A factor above 1 means growth; below 1 means decay.

The value of a car is modeled by $V(t) = 24000(0.88)^t$, where $t$ is the number of years after purchase. Which statement best describes the value of the car?

• (A) It grows by $88\%$ each year.
• (B) It decays by $12\%$ each year. ✓
• (C) It decays by $88\%$ each year.
• (D) It grows by $12\%$ each year.

In $y = a(b)^t$, a base $b$ between $0$ and $1$ represents decay, and the yearly percent rate is $1 - b$.

A savings account starts with $\$2000$ and increases by $5\%$ each year. Which equation models the balance $A$ after $t$ years?

• (A) $A = 2000(0.05)^t$
• (B) $A = 2000(1.5)^t$
• (C) $A = 2000(0.95)^t$
• (D) $A = 2000(1.05)^t$ ✓

For a percent increase, the base is $1 + r$; the initial amount stays as the leading coefficient.

In a geometric sequence, the third term is $25$ and the fifth term is $625$. Which value could be the common ratio?

• (A) $\tfrac{1}{25}$
• (B) $5$ ✓
• (C) $\tfrac{1}{5}$
• (D) $25$

Each step in a geometric sequence multiplies by r. Going from the 3rd term to the 5th is two steps, so the ratio of those terms is r².

A colony of bacteria doubles every hour and starts with $300$ bacteria. The count is modeled by $N(t) = 300(2)^t$, where $t$ is in hours. How many bacteria are present after $4$ hours?

• (A) $1200$
• (B) $9600$
• (C) $4800$ ✓
• (D) $2400$

Evaluate an exponential by computing the power first, then multiplying by the initial amount; doubling is repeated multiplication, not addition.

The value of a rare coin is modeled by $V(t) = 150(1.07)^t$ dollars, where $t$ is the number of years after it was bought. Which statement is correct?

• (A) It was purchased for $\$150$ and increases by $\$7$ each year.
• (B) It was purchased for $\$150$ and increases by $7\%$ each year. ✓
• (C) It was purchased for $\$150$ and increases by $107\%$ each year.
• (D) It was purchased for $\$150$ and decreases by $7\%$ each year.

In $y = a(b)^t$, $a$ is the starting value and $b - 1$ is the percent rate; $b > 1$ indicates growth.

A car purchased for $\$24{,}000$ loses $12\%$ of its value each year. Which function models the value of the car, $V$, after $t$ years?

• (A) $V = 24000(1.12)^t$
• (B) $V = 24000(0.12)^t$
• (C) $V = 24000(0.88)^t$ ✓
• (D) $V = 24000(0.88t)$

A yearly loss of 12% keeps 88% of the value each year, so the base is $1 - 0.12 = 0.88$. The starting value $24{,}000$ is the coefficient and $t$ is the exponent. Using $1.12$ would model growth, and $0.12$ as a base would throw away nearly all the value each year.

A car purchased for $\$20{,}000$ loses $15\%$ of its value each year. Which function models the value of the car, $V(t)$, after $t$ years?

• (A) $V(t) = 20000(0.85)^t$ ✓
• (B) $V(t) = 20000(0.15)^t$
• (C) $V(t) = 20000(1.15)^t$
• (D) $V(t) = 20000 - 0.15t$

Losing $15\%$ each year means the car keeps $85\%$ of its value annually, so the decay factor is $0.85$. Using $1.15$ would model growth, using $0.15$ would mean the car keeps only $15\%$ each year, and the linear expression does not model percent change.

A scientist places $200$ bacteria in a dish. The number of bacteria doubles every hour. Write a function $B(t)$ that models the number of bacteria after $t$ hours. Then determine the number of bacteria in the dish after $5$ hours.

Doubling every hour is exponential growth with growth factor 2 and initial value 200, so the model is $B(t)=200(2)^t$. Evaluating at $t=5$ gives $200\cdot 32 = 6400$ bacteria.

A new car is purchased for $\$25{,}000$. The value of the car decreases by $15\%$ each year. Which function models the value of the car, $V(t)$, after $t$ years?

• (A) $V(t) = 25000(1.15)^t$
• (B) $V(t) = 25000(0.15)^t$
• (C) $V(t) = 25000 - 0.85t$
• (D) $V(t) = 25000(0.85)^t$ ✓

A $15\%$ decrease each year means the car retains $100\% - 15\% = 85\%$ of its value annually, so the decay factor is $0.85$ raised to the power $t$. Using $1.15$ models growth, and using $0.15$ would mean the car loses $85\%$ per year, both common mistakes.

The population of a town is currently 500 people and is increasing by 3% each year. Which equation models the population, $P$, after $t$ years?

• (A) $P = 500(3)^t$
• (B) $P = 500(1.3)^t$
• (C) $P = 500(1.03)^t$ ✓
• (D) $P = 500(0.03)^t$

A 3% yearly increase means each year the population is multiplied by $1 + 0.03 = 1.03$. Using $0.03$ alone would shrink the population, $1.3$ represents a 30% increase, and a base of $3$ would triple the population each year.

A bacteria culture starts with 800 bacteria, and the population doubles every 3 hours. Write a function $P(t)$ that models the number of bacteria after $t$ hours. Then use your function to determine the number of bacteria after 12 hours.

Exponential growth has the form $P(t)=a\cdot b^{t/k}$, where $a$ is the starting amount, $b$ is the growth factor, and $k$ is the time needed for one growth cycle. In 12 hours the population doubles $12\div 3=4$ times, so $800\cdot 2^4=12{,}800$.

The deer population in a nature preserve is modeled by the function $P(t) = 500(1.06)^t$, where $t$ is the number of years since the population was first counted. State the annual percent growth rate of the population, and explain how you know. Then determine, to the nearest whole number, the population predicted by the model after 3 years.

An exponential model $P(t) = a(1 + r)^t$ grows by the same percent each year; since the base is $1.06$, the population increases by $6\%$ annually. Evaluating at $t = 3$ gives $500(1.06)^3 \approx 595.5$, which rounds to 596 deer.

A town's population is currently 500 people and is increasing by 3% each year. Which function models the population, $P(t)$, after $t$ years?

• (A) $P(t) = 500(0.97)^t$
• (B) $P(t) = 500(3)^t$
• (C) $P(t) = 500(1.3)^t$
• (D) $P(t) = 500(1.03)^t$ ✓

A 3% annual increase means the population is multiplied by $1 + 0.03 = 1.03$ each year. Writing 3% as 0.3 gives the incorrect base 1.3, using $0.97$ models decay instead of growth, and a base of 3 would mean tripling every year.

A town's population is currently 500 people and is expected to grow by 6% each year. Which function models the population, $P(t)$, after $t$ years?

• (A) $P(t) = 500(1.06)^t$ ✓
• (B) $P(t) = 500(1.6)^t$
• (C) $P(t) = 500(0.94)^t$
• (D) $P(t) = 500(0.06)^t$

A 6% yearly increase means the population is multiplied by $1 + 0.06 = 1.06$ each year, so the model is $500(1.06)^t$. Using $0.06$ as the base would shrink the population toward zero, $1.6$ represents 60% growth, and $0.94$ models a 6% decay instead of growth.

A new car is purchased for $\$24{,}000$. The value of the car decreases by $15\%$ each year. Write a function $V(t)$ that models the value of the car $t$ years after purchase, and determine the value of the car, to the nearest dollar, 3 years after purchase.

Exponential decay uses the form initial value times (decay factor)$^t$. Losing 15% per year means keeping 85%, so the factor is $0.85$. Evaluating at $t = 3$ gives $24000(0.85)^3 = 14{,}739$ dollars exactly.

Mia deposits $\$3000$ into a savings account that earns $4\%$ interest per year, compounded annually. She makes no other deposits or withdrawals. Write an equation for the amount of money $A$, in dollars, in the account after $t$ years. Then determine, to the nearest cent, the amount of money in the account after 2 years.

Compound interest is modeled by exponential growth, $A = P(1 + r)^t$, because each year the balance is multiplied by the same growth factor of $1.04$. Evaluating the model at $t = 2$ gives $3000(1.0816) = 3244.80$ dollars.

An employee's salary (in thousands of dollars, with $t$ = years after hire) can follow one of four plans: $a(t) = 2^{t} + 25$, $\ b(t) = 10t + 75$, $\ c(t) = \sqrt{400t} + 80$, $\ d(t) = 2(t+1)^2 - 10t + 50$. Which plan pays the most in year $8$?

• (A) $a(t) = 2^{t} + 25$ ✓
• (B) $c(t) = \sqrt{400t} + 80$
• (C) $b(t) = 10t + 75$
• (D) $d(t) = 2(t+1)^2 - 10t + 50$

Exponential growth eventually overtakes linear, quadratic and root models. Evaluate each function at the given year and compare.

The number of members of a club is modeled by $M = 1000(1.21)^t$, where $t$ is time in years. Which expression models the number of members if $x$ represents the number of $6$-month periods?

• (A) $M = 1000(1.1)^x$ ✓
• (B) $M = 1000(0.605)^x$
• (C) $M = 1000(1.21)^{2x}$
• (D) $M = 1000(1.105)^x$

Changing the time unit rescales the exponent; use the power rule so the new base is a root of the old base, keeping the same actual growth.

A radioactive sample has a half-life of $5$ years, and a sample starts with $80$ grams. Which equation models the mass $m$, in grams, remaining after $t$ years?

• (A) $m = 80(2)^{t/5}$
• (B) $m = 80\left(\dfrac{1}{2}\right)^{5t}$
• (C) $m = 80\left(\dfrac{1}{2}\right)^{t/5}$ ✓
• (D) $m = 80\left(\dfrac{1}{2}\right)^{t}$

For half-life decay the base is $\tfrac{1}{2}$ and the exponent is time divided by the length of one half-life.

Marisol has $2000$ dollars to invest and is comparing two options. Option A adds a fixed $120$ dollars to the account at the end of each year. Option B pays $5\%$ interest compounded annually. Write a function $A(t)$ for the value of Option A and a function $B(t)$ for the value of Option B, where $t$ is the number of years. Then determine the value of each option after $10$ years, rounded to the nearest dollar, and state which option is worth more after $10$ years and by how much.

Option A adds the same amount yearly, giving a linear model, while Option B multiplies by a fixed growth factor of $1.05$ each year, giving an exponential model. Evaluating each function at $t = 10$ and comparing shows the compounding of Option B overtakes the steady additions of Option A, making Option B worth about $58$ dollars more after $10$ years.

A car is purchased for $\$24{,}000$ and loses $15\%$ of its value each year. Which function represents the value of the car, in dollars, after $t$ years?

• (A) $24000(0.15)^t$
• (B) $24000(1.15)^t$
• (C) $24000(0.85t)$
• (D) $24000(0.85)^t$ ✓

Losing $15\%$ each year means the car keeps $85\%$ of its value, so the base is $0.85$. The repeated multiplication over $t$ years makes it exponential: $24000(0.85)^t$.

The value of an investment is modeled by $V(t) = 800(1.05)^t$, where $V(t)$ is the value in dollars and $t$ is the time in years. State the initial value of the investment and the annual growth rate as a percent. Then determine the value of the investment after $2$ years, rounded to the nearest cent.

In an exponential model $a\,b^t$, the constant $a$ is the initial amount and $b = 1 + r$ gives the rate $r$. Here $a = 800$ and $b = 1.05$ correspond to a 5% annual increase, and evaluating at $t=2$ gives the value after two years.

A new car is purchased for 28,000 dollars. The value of the car decreases by 12% each year. Write an exponential function $V(t)$ that models the value of the car, in dollars, after $t$ years. Use your function to determine the value of the car, to the nearest dollar, after 5 years. Then determine whether the car will be worth less than 10,000 dollars after 9 years, and justify your answer.

A constant 12% yearly loss is exponential decay with base $0.88$, giving $V(t) = 28000(0.88)^t$. Evaluating at $t = 5$ yields about 14,776 dollars, and at $t = 9$ yields about 8,861 dollars. Since 8,861 is less than 10,000, the car is worth less than 10,000 dollars after 9 years.