Question

Suppose that $20,000 is invested in an account for which interest is compounding continuously at 3.14%. What is the value after 5 years? After 10 years? (Round your answers to two decimal places.) After how many years will the original investment be doubled? (Round your answer to two decimal places.)

Answer 1

Answer:

value after 5 years = $23,399.91

after 10 years = $27,377.79

Time it takes for the amount to double = 22.07 years

Step-by-step explanation:

For amounts that are compounded continuously, it means that the interest rate is is added to the investment amount at an infinite number of time, and the formula is given as:

A = P $e^{r.t}$, where:

A = Future value

P = present value

e = constant ≈ 2.7183

r = interest rate in decimal form

t = years

Now for value after 5 years;

A = ???

P = $20,000

r = 3.14% = 0.0314

t =  5 years

∴ A = P $e^{r.t}$

= 20,000 $e^{0.0314*5}$

= 20,000 × $e^{0.157}$ = 20,000 × 1.169995 = $23,399.91 ( to 2 decimal places)

(Note that the function '$e$' can be punched directly from the calculator)

value after 10 years;

A = ???

P = $20,000

r = 3.14% = 0.0314

t =  10 years

∴ A = P$e^{r.t}$

= 20,000 × $e^{0.0314 * 10}$

= 20,000 × $e^{0.314}$ = $27,377.79 ( to 2 decimal places)

Time it will take to double the original investment;

A = P $e^{r.t}$

where;

A = 40,000

P = 20,000

r = 0.0314

t =???

40,000 = 20,000 × $e^{0.0314 * t}$

$\frac{40,000}{20,000} = \frac{20,000}{20,000} * e^{0.0314*t}$ (divide both sides by 20,000)

2 = $e^{0.0314 * t}$

Next take the natural logarithm of both sides

㏑(2) = ㏑$e^{0.0314 *t}$      (㏑$e$ = 1; and the exponent can be brought down )

= 0.6931 = 0.0314 × t × 1

∴ t = $\frac{0.06931}{0.0314}$ = 22.07 years ( to 2 decimal places)