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Exponential Function Solving - Decay (Discrete, Mis-matched Time Units) Scenario to Starting Value

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 - Decay (Discrete, Mis-matched Time Units) Scenario to Starting Value Worksheet

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Exponential Function Solving - Decay (Discrete, Mis-matched Time Unit...
1
A toxin starts at a certain concentration. Each weekly dialysis reduces it by 7%. After 28 days it has decreased to a concentration of 65mg/L.
How would you solve for the starting concentration given this scenario?
a A LaTex expression showing C sub 0 = C times (1-r) to the power of t times 7
b A LaTex expression showing C sub 0 = C over (1-r) to the power of \frac{t {7 }}
c A LaTex expression showing C sub 0 = C over (1+r) to the power of \frac{t {7 }}
2
A charitable endowment starts with a certain amount of money. Each yearly it disburses 7% of its remaining funds. After 1460 days its funds have decreased to $0.
How would you solve for the starting cash given this scenario?
a A LaTex expression showing P sub 0 = P over (1+r) to the power of \frac{t {365 }}
b A LaTex expression showing P sub 0 = P over (1-r) to the power of \frac{t {365 }}
c A LaTex expression showing P sub 0 = P times (1-r) to the power of t times 365
3
A toxin starts at a certain concentration. Each weekly dialysis reduces it by 7%. After 63 days it has decreased to a concentration of 4mg/L.
How would you solve for the starting concentration given this scenario?
a A LaTex expression showing C sub 0 = C times (1-r) to the power of t times 7
b A LaTex expression showing C sub 0 = C over (1+r) to the power of \frac{t {7 }}
c A LaTex expression showing C sub 0 = C over (1-r) to the power of \frac{t {7 }}
4
A toxin starts at a certain concentration. Each weekly dialysis reduces it by 9%. After 56 days it has decreased to a concentration of 3mg/L.
How would you solve for the starting concentration given this scenario?
a A LaTex expression showing C sub 0 = C over (1+r) to the power of \frac{t {7 }}
b A LaTex expression showing C sub 0 = C times (1-r) to the power of t times 7
c A LaTex expression showing C sub 0 = C over (1-r) to the power of \frac{t {7 }}
5
A charitable endowment starts with a certain amount of money. Each daily it disburses 4% of its remaining funds. After 2 years its funds have decreased to $276.
How would you solve for the starting cash given this scenario?
a A LaTex expression showing P sub 0 = P over (1-r) to the power of t times 365
b A LaTex expression showing P sub 0 = P over (1+r) to the power of t times 365
c A LaTex expression showing P sub 0 = P times (1-r) to the power of t over 365
6
A charitable endowment starts with a certain amount of money. Each weekly it disburses 3% of its remaining funds. After 49 days its funds have decreased to $112.
How would you solve for the starting cash given this scenario?
a A LaTex expression showing P sub 0 = P over (1-r) to the power of \frac{t {7 }}
b A LaTex expression showing P sub 0 = P times (1-r) to the power of t times 7
c A LaTex expression showing P sub 0 = P over (1+r) to the power of \frac{t {7 }}
7
A charitable endowment starts with a certain amount of money. Each daily it disburses 9% of its remaining funds. After 6 weeks its funds have decreased to $283.
How would you solve for the starting cash given this scenario?
a A LaTex expression showing P sub 0 = P over (1+r) to the power of t times 7
b A LaTex expression showing P sub 0 = P over (1-r) to the power of t times 7
c A LaTex expression showing P sub 0 = P times (1-r) to the power of t over 7
8
A charitable endowment starts with a certain amount of money. Each yearly it disburses 3% of its remaining funds. After 730 days its funds have decreased to $0.
How would you solve for the starting cash given this scenario?
a A LaTex expression showing P sub 0 = P over (1-r) to the power of \frac{t {365 }}
b A LaTex expression showing P sub 0 = P over (1+r) to the power of \frac{t {365 }}