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Exponential Function Solving - Growth (Discrete) Scenario to Time

Level 1

This math topic focuses on solving real-world growth scenarios using exponential functions. Problems involve calculating the time required for populations, savings accounts, and credit card debts to grow or reach a certain target based on a fixed percent growth or interest rate. The calculations require understanding of exponential growth formulas, specifically manipulating equations to isolate and solve for the variable representing time. Each problem presents a context (e.g., rabbit population growth, savings interest accrual) and asks to determine the time span needed to achieve a specified result, using the exponential growth model.

Work on practice problems directly here, or download the printable pdf worksheet to practice offline.

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Exponential Function Solving - Growth (Discrete) Scenario to Time Worksheet

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Exponential Function Solving - Growth (Discrete) Scenario to Time
1
A rabbit population starts at 200. Each subsequent yearly breeding season it grows by 4%. After a certain number of years it has increased to a population of 243 rabbits.
Solve for the time given this scenario?
a A LaTex expression showing t = \ln{\frac{P over P sub 0 }}{\ln{(1+r)}}
b A LaTex expression showing 8 + t = \frac{\ln{P times P sub 0 }}{\ln{(1+r)}}
c A LaTex expression showing 5 + t = \frac{\ln{P times P sub 0 }}{\ln{(1+r)}}
2
A savings account starts with $500. Each subsequent month it earns 9% in interest. After a certain number of months it has $914.
Solve for the time given this scenario?
a A LaTex expression showing 4 + t = \ln{\frac{P over P sub 0 }}{\ln{(1-r)}}
b A LaTex expression showing t = \ln{\frac{P over P sub 0 }}{\ln{(1+r)}}
c A LaTex expression showing 6 + t = \ln{\frac{P over P sub 0 }}{\ln{(1-r)}}
d A LaTex expression showing 5 + t = \frac{\ln{P times P sub 0 }}{\ln{(1+r)}}
3
An insect population starts at 900. Each subsequent yearly breeding season it grows by 7%. After a certain number of years it has increased to a population of 1,179.
Solve for the time given this scenario?
a A LaTex expression showing t = \ln{\frac{P over P sub 0 }}{\ln{(1+r)}}
b A LaTex expression showing 4 + t = \ln{\frac{P over P sub 0 }}{\ln{(1-r)}}
c A LaTex expression showing 9 + t = \ln{\frac{P over P sub 0 }}{\ln{(1-r)}}
d A LaTex expression showing 3 + t = \ln{\frac{P over P sub 0 }}{\ln{(1-r)}}
4
A savings account starts with $600. Each subsequent year it earns 8% in interest. After a certain number of years it has $1,028.
Solve for the time given this scenario?
a A LaTex expression showing 4 + t = \ln{\frac{P over P sub 0 }}{\ln{(1-r)}}
b A LaTex expression showing 3 + t = \ln{\frac{P over P sub 0 }}{\ln{(1-r)}}
c A LaTex expression showing 5 + t = \ln{\frac{P over P sub 0 }}{\ln{(1-r)}}
d A LaTex expression showing t = \ln{\frac{P over P sub 0 }}{\ln{(1+r)}}
5
A credit card starts with $300 of debt. Each subsequent month it grows by 9% in interest. After a certain number of months the debt has grown to $503.
Solve for the time given this scenario?
a A LaTex expression showing 6 + t = \ln{\frac{D over D sub 0 }}{\ln{(1-r)}}
b A LaTex expression showing 2 + t = \ln{\frac{D over D sub 0 }}{\ln{(1-r)}}
c A LaTex expression showing t = \ln{\frac{D over D sub 0 }}{\ln{(1+r)}}
d A LaTex expression showing 4 + t = \frac{\ln{D times D sub 0 }}{\ln{(1+r)}}
6
A credit card starts with $600 of debt. Each subsequent quarter it grows by 7% in interest. After a certain number of quarters the debt has grown to $735.
Solve for the time given this scenario?
a A LaTex expression showing 9 + t = \frac{\ln{D times D sub 0 }}{\ln{(1+r)}}
b A LaTex expression showing 6 + t = \ln{\frac{D over D sub 0 }}{\ln{(1-r)}}
c A LaTex expression showing t = \ln{\frac{D over D sub 0 }}{\ln{(1+r)}}
d A LaTex expression showing 6 + t = \frac{\ln{D times D sub 0 }}{\ln{(1+r)}}
7
A credit card starts with $700 of debt. Each subsequent month it grows by 9% in interest. After a certain number of months the debt has grown to $906.
Solve for the time given this scenario?
a A LaTex expression showing 9 + t = \ln{\frac{D over D sub 0 }}{\ln{(1-r)}}
b A LaTex expression showing t = \ln{\frac{D over D sub 0 }}{\ln{(1+r)}}
c A LaTex expression showing 9 + t = \frac{\ln{D times D sub 0 }}{\ln{(1+r)}}
d A LaTex expression showing 3 + t = \ln{\frac{D over D sub 0 }}{\ln{(1-r)}}
8
A savings account starts with $500. Each subsequent quarter it earns 9% in interest. After a certain number of quarters it has $838.
Solve for the time given this scenario?
a A LaTex expression showing 1 + t = \frac{\ln{P times P sub 0 }}{\ln{(1+r)}}
b A LaTex expression showing t = \ln{\frac{P over P sub 0 }}{\ln{(1+r)}}
c A LaTex expression showing 2 + t = \frac{\ln{P times P sub 0 }}{\ln{(1+r)}}
d A LaTex expression showing 6 + t = \frac{\ln{P times P sub 0 }}{\ln{(1+r)}}