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Exponential Function Solution Equation - Growth (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.

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Exponential Function Solution Equation - Growth (Discrete, Mis-matched Time Units) Scenario to Starting Value Worksheet

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Exponential Function Solution Equation - Growth (Discrete, Mis-matche...
1
A credit card starts with a certain amount of debt. Each subsequent quarter it grows by 6% in interest. After 24 months the debt has grown to $809.
Rearrange the exponential equation to solve for for the starting debt given this scenario?
a A LaTex expression showing D sub 0 = 809 over (1-0.06) to the power of \frac{24 {3 }}
b A LaTex expression showing D sub 0 = 809 over (1+0.06) to the power of \frac{24 {3 }}
c A LaTex expression showing D sub 0 = 809 times (1+0.06) to the power of 24 times 3
2
A savings account starts with a certain amount of cash. Each subsequent month it earns 9% in interest. After 6 years it has $838.
Rearrange the exponential equation to solve for for the starting cash given this scenario?
a A LaTex expression showing P sub 0 = 838 over (1+0.09) to the power of 6 times 12
b A LaTex expression showing P sub 0 = 838 times (1+0.09) to the power of 6 over 12
c A LaTex expression showing P sub 0 = 838 over (1-0.09) to the power of 6 times 12
3
A credit card starts with a certain amount of debt. Each subsequent month it grows by 5% in interest. After 8 quarters the debt has grown to $295.
Rearrange the exponential equation to solve for for the starting debt given this scenario?
a A LaTex expression showing D sub 0 = 295 times (1+0.05) to the power of 8 over 3
b A LaTex expression showing D sub 0 = 295 over (1+0.05) to the power of 8 times 3
c A LaTex expression showing D sub 0 = 295 over (1-0.05) to the power of 8 times 3
4
A savings account starts with a certain amount of cash. Each subsequent quarter it earns 3% in interest. After 6 years it has $835.
Rearrange the exponential equation to solve for for the starting cash given this scenario?
a A LaTex expression showing P sub 0 = 835 over (1+0.03) to the power of 6 times 4
b A LaTex expression showing P sub 0 = 835 times (1+0.03) to the power of 6 over 4
5
A credit card starts with a certain amount of debt. Each subsequent quarter it grows by 8% in interest. After 2 years the debt has grown to $1,049.
Rearrange the exponential equation to solve for for the starting debt given this scenario?
a A LaTex expression showing D sub 0 = 1049 over (1+0.08) to the power of 2 times 4
b A LaTex expression showing D sub 0 = 1049 over (1-0.08) to the power of 2 times 4
c A LaTex expression showing D sub 0 = 1049 times (1+0.08) to the power of 2 over 4
6
A credit card starts with a certain amount of debt. Each subsequent quarter it grows by 2% in interest. After 5 years the debt has grown to $331.
Rearrange the exponential equation to solve for for the starting debt given this scenario?
a A LaTex expression showing D sub 0 = 331 over (1-0.02) to the power of 5 times 4
b A LaTex expression showing D sub 0 = 331 over (1+0.02) to the power of 5 times 4
7
A savings account starts with a certain amount of cash. Each subsequent year it earns 9% in interest. After 36 months it has $8,900.
Rearrange the exponential equation to solve for for the starting cash given this scenario?
a A LaTex expression showing P sub 0 = 8900 times (1+0.09) to the power of 36 times 12
b A LaTex expression showing P sub 0 = 8900 over (1-0.09) to the power of \frac{36 {12 }}
c A LaTex expression showing P sub 0 = 8900 over (1+0.09) to the power of \frac{36 {12 }}
8
A savings account starts with a certain amount of cash. Each subsequent year it earns 2% in interest. After 32 quarters it has $1,130.
Rearrange the exponential equation to solve for for the starting cash given this scenario?
a A LaTex expression showing P sub 0 = 1130 over (1+0.02) to the power of \frac{32 {4 }}
b A LaTex expression showing P sub 0 = 1130 over (1-0.02) to the power of \frac{32 {4 }}
c A LaTex expression showing P sub 0 = 1130 times (1+0.02) to the power of 32 times 4