Exponential Growth and Decay: Graphing & Logarithms in Action

A population of bacteria grows exponentially according to the model P(t) = P0 * e^(kt). If a culture starts with 500 bacteria and doubles every 3 hours, what is the growth constant k?

• 0.154151

• 0.693147

• 0.115524

• 0.231049

Tags

CCSS.HSF.LE.A.4

Using the same bacteria growth model, how many bacteria will there be after 12 hours? Graph the function P(t) = 500 * e^(kt) using the value of k you found in the previous question.

• P(12) = 500 * e^(6k)

• P(12) = 500 * e^(12k)

• P(12) = 500 * e^(24k)

• P(12) = 1000 * e^(12k)

A certain investment grows continuously at a rate of 5% per year. If you invest $1000, how much will the investment be worth after 10 years? Use the formula A = Pe^(rt) and graph the function A(t) = 1000 * e^(0.05t).

• 1648.72

• 1200.50

• 2000.00

• 1500.00

If the value of an investment is modeled by the function V(t) = 2000 * e^(0.03t), how long will it take for the investment to reach $3000? Use logarithms to solve for t.

• 18.3

• 15.2

• 20.5

• 25.0

Tags

CCSS.HSF.LE.A.4

A car depreciates in value according to the model V(t) = V0 * e^(-kt). If a car is initially worth $20,000 and loses 15% of its value each year, what is the value of k?

• 0.1625

• 0.200

• 0.125

• 0.075

Using the depreciation model from the previous question, what will the value of the car be after 5 years? Graph the function V(t) = 20000 * e^(-0.15t).

• 9448.00

• 12000.00

• 7500.00

• 11000.00

A certain radioactive substance has a half-life of 5 years. If you start with 80 grams, how much will remain after 15 years? Use the exponential decay model N(t) = N0 * e^(-kt) and find k using the half-life information.

• 40 grams

• 20 grams

• 10 grams

• 5 grams

Tags

CCSS.HSF.LE.A.4

Using the decay model from the previous question, graph the function N(t) = 80 * e^(-kt) and determine the time it takes for the substance to decay to 10 grams.

• t = -ln(10) / k

• t = 80 / k

• t = -ln(2) / k

• t = -ln(1/8) / k

A researcher finds that the number of a certain species of fish in a lake can be modeled by the equation N(t) = 1000 * e^(0.1t). How many fish will there be after 10 years?

• 2718

• 1500

• 3200

• 2000

If the fish population reaches 2000, how long will it take? Use logarithms to solve for t and graph the function N(t) = 1000 * e^(0.1t).

• 8.25

• 6.93

• 5.00

• 10.50

Tags

CCSS.HSF.LE.A.4