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Exponential Function Solving - Growth (Continuous) - Scenario to Time

Level 1

This math topic covers solving for time in continuous growth scenarios using exponential functions. It provides practice in understanding and calculating the time required for different entities (such as savings accounts, rabbit populations, app downloads, bacterial growth, and debt growth) to reach a certain growth under continuous percentage increases over specified periods. Each problem presents a real-life situation where learners apply logarithms and exponential calculations to find how long it takes for an initial amount to grow to a predefined amount. This includes manipulating exponential equations to isolate and solve for the variable representing time.

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

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

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Exponential Function Solving - Growth (Continuous) - Scenario to Time
1
A savings account starts with $300. It grows continuously at 5% interest per month. After a certain number of months it has $404.
Solve for the time given this scenario?
a A LaTex expression showing t = +\ln{\frac{P over P sub 0 }}{r}
b A LaTex expression showing 4 + t = +r over \ln{\frac{P {P sub 0 }}}
c A LaTex expression showing 0 + t = +r over \ln{\frac{P {P sub 0 }}}
d A LaTex expression showing 1 + t = +r over \ln{\frac{P {P sub 0 }}}
2
A rabbit population starts at 500. It grows continuously at 6% growth per quarter. After a certain number of quarters it has increased to a population of 858 rabbits.
Solve for the time given this scenario?
a A LaTex expression showing t = +\ln{\frac{P over P sub 0 }}{r}
b A LaTex expression showing 9 + t = +\frac{\ln{P times P sub 0 }}{r}
c A LaTex expression showing 7 + t = +\frac{\ln{P times P sub 0 }}{r}
d A LaTex expression showing 5 + t = +r over \ln{\frac{P {P sub 0 }}}
3
An app starts with 200 downloads. Its download count grows continually by 7% each week.After a certain number of weeks it has 246 downloads.
Solve for the time given this scenario?
a A LaTex expression showing t = +\ln{\frac{A over A sub 0 }}{r}
b A LaTex expression showing 4 + t = +\frac{\ln{A times A sub 0 }}{r}
c A LaTex expression showing 2 + t = +r over \ln{\frac{A {A sub 0 }}}
d A LaTex expression showing 5 + t = +\frac{\ln{A times A sub 0 }}{r}
4
A bacteria population starts at 600. It grows continuously at 4% growth per year. After a certain number of years it has increased to a population of 793.
Solve for the time given this scenario?
a A LaTex expression showing 8 + t = +r over \ln{\frac{P {P sub 0 }}}
b A LaTex expression showing t = +\ln{\frac{P over P sub 0 }}{r}
c A LaTex expression showing 7 + t = +\frac{\ln{P times P sub 0 }}{r}
d A LaTex expression showing 3 + t = +\frac{\ln{P times P sub 0 }}{r}
5
A credit card starts with $400 of debt. It grows continuously at 6% interest per year. After a certain number of years the debt has grown to $478.
Solve for the time given this scenario?
a A LaTex expression showing 9 + t = +r over \ln{\frac{D {D sub 0 }}}
b A LaTex expression showing 6 + t = +r over \ln{\frac{D {D sub 0 }}}
c A LaTex expression showing t = +\ln{\frac{D over D sub 0 }}{r}
6
A savings account starts with $200. It grows continuously at 6% interest per year. After a certain number of years it has $269.
Solve for the time given this scenario?
a A LaTex expression showing 6 + t = +\frac{\ln{P times P sub 0 }}{r}
b A LaTex expression showing t = +\ln{\frac{P over P sub 0 }}{r}
c A LaTex expression showing 1 + t = +r over \ln{\frac{P {P sub 0 }}}
7
A savings account starts with $600. It grows continuously at 8% interest per year. After a certain number of years it has $1,232.
Solve for the time given this scenario?
a A LaTex expression showing 2 + t = +r over \ln{\frac{P {P sub 0 }}}
b A LaTex expression showing 9 + t = +\frac{\ln{P times P sub 0 }}{r}
c A LaTex expression showing t = +\ln{\frac{P over P sub 0 }}{r}
d A LaTex expression showing 0 + t = +\frac{\ln{P times P sub 0 }}{r}
8
A company's share price starts at $600. It grows continuously at 3% growth per quarter. After a certain number of quarters it has a share price of $697.
Solve for the time given this scenario?
a A LaTex expression showing 1 + t = +\frac{\ln{S times S sub 0 }}{r}
b A LaTex expression showing t = +\ln{\frac{S over S sub 0 }}{r}
c A LaTex expression showing 0 + t = +r over \ln{\frac{S {S sub 0 }}}
d A LaTex expression showing 8 + t = +\frac{\ln{S times S sub 0 }}{r}