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Exponential Function Solving - Growth (Continuous, Mis-matched Time Units) - Scenario to Time

Level 1

The topics in this unit focus on mastering exponential growth and decay functions. Work on practice problems directly here, or download the printable pdf worksheet to practice offline.

WorksheetView

Exponential Function Solving - Growth (Continuous, Mis-matched Time Units) - Scenario to Time Worksheet

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Exponential Function Solving - Growth (Continuous, Mis-matched Time U...
1
An insect population starts at 600. It grows continuously at 7% growth per day. After a certain number of weeks it has increased to a population of 1,126.
How would you solve for the time given this scenario?
a A LaTex expression showing t = +7 times \ln{\frac{P over P sub 0 }}{r}
b A LaTex expression showing t = +7 times \frac{\ln{P times P sub 0 }}{r}
c A LaTex expression showing t = +1 over 7 times \ln{\frac{P over P sub 0 }}{r}
2
A company's share price starts at $300. It grows continuously at 6% growth per year. After a certain number of months it has a share price of $381.
How would you solve for the time given this scenario?
a A LaTex expression showing t = +1 over 12 times \frac{\ln{S times S sub 0 }}{r}
b A LaTex expression showing t = +12 times r over \ln{\frac{S {S sub 0 }}}
c A LaTex expression showing t = +12 times \ln{\frac{S over S sub 0 }}{r}
3
A rabbit population starts at 400. It grows continuously at 7% growth per year. After a certain number of quarters it has increased to a population of 567 rabbits.
How would you solve for the time given this scenario?
a A LaTex expression showing t = +1 over 4 times \frac{\ln{P times P sub 0 }}{r}
b A LaTex expression showing t = +4 times \ln{\frac{P over P sub 0 }}{r}
c A LaTex expression showing t = +1 over 4 times \ln{\frac{P over P sub 0 }}{r}
4
An insect population starts at 200. It grows continuously at 7% growth per day. After a certain number of years it has increased to a population of 350.
How would you solve for the time given this scenario?
a A LaTex expression showing t = +1 over 365 times r over \ln{\frac{P {P sub 0 }}}
b A LaTex expression showing t = +1 over 365 times \ln{\frac{P over P sub 0 }}{r}
c A LaTex expression showing t = +365 times \frac{\ln{P times P sub 0 }}{r}
5
A social media post starts with 900 views. Its view count grows continually by 7% each day.After a certain number of years it has 1,110 views.
How would you solve for the time given this scenario?
a A LaTex expression showing t = +1 over 365 times r over \ln{\frac{V {V sub 0 }}}
b A LaTex expression showing t = +1 over 365 times \ln{\frac{V over V sub 0 }}{r}
c A LaTex expression showing t = +365 times \ln{\frac{V over V sub 0 }}{r}
6
A social media post starts with 500 views. Its view count grows continually by 4% each month.After a certain number of years it has 563 views.
How would you solve for the time given this scenario?
a A LaTex expression showing t = +12 times \ln{\frac{V over V sub 0 }}{r}
b A LaTex expression showing t = +1 over 12 times r over \ln{\frac{V {V sub 0 }}}
c A LaTex expression showing t = +1 over 12 times \ln{\frac{V over V sub 0 }}{r}
7
A credit card starts with $400 of debt. It grows continuously at 6% interest per quarter. After a certain number of months the debt has grown to $539.
How would you solve for the time given this scenario?
a A LaTex expression showing t = +1 over 3 times \frac{\ln{D times D sub 0 }}{r}
b A LaTex expression showing t = +1 over 3 times \ln{\frac{D over D sub 0 }}{r}
c A LaTex expression showing t = +3 times \ln{\frac{D over D sub 0 }}{r}
8
A social media post starts with 800 views. Its view count grows continually by 7% each year.After a certain number of months it has 920 views.
How would you solve for the time given this scenario?
a A LaTex expression showing t = +1 over 12 times \frac{\ln{V times V sub 0 }}{r}
b A LaTex expression showing t = +12 times r over \ln{\frac{V {V sub 0 }}}
c A LaTex expression showing t = +12 times \ln{\frac{V over V sub 0 }}{r}