Exponential Growth and Decay worksheet

1. Time to reach the target (periods): 4.47

1. Smooth, nonstop growth, so the rate k is positive

   $k$$=0.06$

2. Divide the target by the starting amount to see how many times bigger it must get

   $\frac{\mathrm{1,700}}{\mathrm{1,300}}$$=1.307692$

3. Logs undo powers, so the natural log of that factor divided by k gives the time t

   $t$$=\frac{\text{ln}\left(1.307692\right)}{0.06}$$=4.471066$

4. Doubling time: the time to double is ln 2 (about 0.693) divided by the continuous rate

   $\mathrm{t₂}$$=\frac{\text{ln}\left(2\right)}{0.06}$$=11.552453$

2. Amount after 9 periods: 1,387.73

1. Gaining 4% each period is the same as multiplying by b = 1 + r

   $b$$=1+0.04$$=1.04$

2. Multiply the starting amount by the factor once for each period

   $P$$=975\times {1.04}^9$$=\mathrm{1,387.729017}$

3. Doubling time: the time to double is ln 2 (about 0.693) divided by the continuous rate

   $\mathrm{t₂}$$=\frac{\text{ln}\left(2\right)}{0.03922071}$$=17.672988$

3. Growth rate per period: 2.51%

1. Divide the final amount by the initial amount to get the overall growth factor

   $\frac{\mathrm{1,500}}{\mathrm{1,200}}$$=1.25$

2. Undo the repeated multiplying: take the t-th root to get the factor for one period (b)

   $b$$={1.25}^{\frac{1}{9}}$$=1.02510365$

3. Take away 1 to leave just the change, then show it as a percent (r)

   $r$$=\left(1.02510365-1\right)\times 100$$=\mathrm{2.510365\%}$

4. Doubling time: the time to double is ln 2 (about 0.693) divided by the continuous rate

   $\mathrm{t₂}$$=\frac{\text{ln}\left(2\right)}{0.02479373}$$=27.956553$

4. Time to reach the target (periods): 16.94

1. Gaining 6% each period is the same as multiplying by b = 1 + r

   $b$$=1+0.06$$=1.06$

2. Divide the target by the starting amount to see how many times bigger it must get

   $\frac{\mathrm{2,750}}{\mathrm{1,025}}$$=2.682927$

3. Logs undo powers, so the log of that factor divided by the log of b counts the periods t

   $t$$=\frac{\text{ln}\left(2.682927\right)}{\text{ln}\left(1.06\right)}$$=16.937134$

4. Doubling time: the time to double is ln 2 (about 0.693) divided by the continuous rate

   $\mathrm{t₂}$$=\frac{\text{ln}\left(2\right)}{0.05826891}$$=11.895661$

5. Continuous growth rate: 7.26%

1. Divide the final amount by the initial amount to get the overall growth factor

   $\frac{\mathrm{2,000}}{900}$$=2.222222$

2. For smooth, nonstop change, the rate k is the natural log of that factor, split over the t periods

   $k$$=\frac{\text{ln}\left(2.222222\right)}{11}$$=0.07259161$

3. Multiply by 100 to show it as a percent

   $k$$=\mathrm{7.259161\%}$

4. Doubling time: the time to double is ln 2 (about 0.693) divided by the continuous rate

   $\mathrm{t₂}$$=\frac{\text{ln}\left(2\right)}{0.07259161}$$=9.548585$

6. Amount after 12 periods: 432.29

1. Losing 5% each period is the same as multiplying by b = 1 − r

   $b$$=1-0.05$$=0.95$

2. Multiply the starting amount by the factor once for each period

   $P$$=800\times {0.95}^{12}$$=432.28807$

3. Half-life: the time to shrink by half is ln 2 (about 0.693) divided by the size of the continuous rate

   $\mathrm{t½}$$=\frac{\text{ln}\left(2\right)}{0.05129329}$$=13.513407$

7. Amount after 10 periods: 700.55

1. Losing 3% each period is the same as multiplying by b = 1 − r

   $b$$=1-0.03$$=0.97$

2. Multiply the starting amount by the factor once for each period

   $P$$=950\times {0.97}^{10}$$=700.552921$

3. Half-life: the time to shrink by half is ln 2 (about 0.693) divided by the size of the continuous rate

   $\mathrm{t½}$$=\frac{\text{ln}\left(2\right)}{0.03045921}$$=22.756573$

8. Time to reach the target (periods): 19.02

1. Smooth, nonstop growth, so the rate k is positive

   $k$$=0.03$

2. Divide the target by the starting amount to see how many times bigger it must get

   $\frac{\mathrm{2,300}}{\mathrm{1,300}}$$=1.769231$

3. Logs undo powers, so the natural log of that factor divided by k gives the time t

   $t$$=\frac{\text{ln}\left(1.769231\right)}{0.03}$$=19.018162$

4. Doubling time: the time to double is ln 2 (about 0.693) divided by the continuous rate

   $\mathrm{t₂}$$=\frac{\text{ln}\left(2\right)}{0.03}$$=23.104906$

9. Amount after 12 periods: 480.21

1. Smooth, nonstop decay, so the rate k is negative

   $k$$=\mathrm{-0.05}$

2. Multiply the starting amount by e (about 2.718) raised to k × t

   $P$$=875\times e^{\left(\mathrm{-0.05}\times 12\right)}$$=480.210182$

3. Half-life: the time to shrink by half is ln 2 (about 0.693) divided by the size of the continuous rate

   $\mathrm{t½}$$=\frac{\text{ln}\left(2\right)}{0.05}$$=13.862944$

10. Growth rate per period: 5.95%

1. Divide the final amount by the initial amount to get the overall growth factor

   $\frac{\mathrm{1,600}}{800}$$=2$

2. Undo the repeated multiplying: take the t-th root to get the factor for one period (b)

   $b$$=2^{\frac{1}{12}}$$=1.05946309$

3. Take away 1 to leave just the change, then show it as a percent (r)

   $r$$=\left(1.05946309-1\right)\times 100$$=\mathrm{5.946309\%}$

4. Doubling time: the time to double is ln 2 (about 0.693) divided by the continuous rate

   $\mathrm{t₂}$$=\frac{\text{ln}\left(2\right)}{0.05776227}$$=12$