you deposit money into a savings account that earns interest. the balance a of the account after t years is…

you deposit money into a savings account that earns interest. the balance a of the account after t years is given by a = 120×1.05^t dollars. (a) make a table that shows the account balance for years 0 through 10. (round your answers to one decimal place.) (b) you want to buy a home - entertainment center that costs $1500. how long will you need to wait to have enough money in your savings account? report your
Answer
Explanation:
Step1: Calculate balance for each year
For (t = 0), (A=120\times1.05^{0}=120\times1 = 120.0) For (t = 1), (A = 120\times1.05^{1}=120\times1.05=126.0) For (t = 2), (A = 120\times1.05^{2}=120\times1.1025 = 132.3) For (t = 3), (A = 120\times1.05^{3}=120\times1.157625=138.9) For (t = 4), (A = 120\times1.05^{4}=120\times1.21550625 = 145.9) For (t = 5), (A = 120\times1.05^{5}=120\times1.2762815625=153.2) For (t = 6), (A = 120\times1.05^{6}=120\times1.340095640625 = 160.8) For (t = 7), (A = 120\times1.05^{7}=120\times1.40710042265625=168.9) For (t = 8), (A = 120\times1.05^{8}=120\times1.4774554437890625 = 177.3) For (t = 9), (A = 120\times1.05^{9}=120\times1.551328215978515625=186.2) For (t = 10), (A = 120\times1.05^{10}=120\times1.6288946267771069375=195.5)
Step2: Solve for (t) when (A = 1500)
Set (1500=120\times1.05^{t}), then (\frac{1500}{120}=1.05^{t}), so (12.5 = 1.05^{t}). Take the natural - logarithm of both sides: (\ln(12.5)=\ln(1.05^{t})). Using the property (\ln(a^{b})=b\ln(a)), we get (\ln(12.5)=t\ln(1.05)). So (t=\frac{\ln(12.5)}{\ln(1.05)}\approx\frac{2.5257}{0.0488}\approx51.8)
| (t) (years) | (A) (dollars) |
|---|---|
| 0 | 120.0 |
| 1 | 126.0 |
| 2 | 132.3 |
| 3 | 138.9 |
| 4 | 145.9 |
| 5 | 153.2 |
| 6 | 160.8 |
| 7 | 168.9 |
| 8 | 177.3 |
| 9 | 186.2 |
| 10 | 195.5 |
Answer:
| (t) (years) | (A) (dollars) |
|---|---|
| 0 | 120.0 |
| 1 | 126.0 |
| 2 | 132.3 |
| 3 | 138.9 |
| 4 | 145.9 |
| 5 | 153.2 |
| 6 | 160.8 |
| 7 | 168.9 |
| 8 | 177.3 |
| 9 | 186.2 |
| 10 | 195.5 |
| (b) Approximately 51.8 years |