Question

If Tanisha has ​$1000 to invest at 7​% per annum compounded semiannually​, how long will it be before she has ​$1600​? If the compounding is​ continuous, how long will it​ be?

Answer 1

Answer:

Using continuous interest 6.83 years before she has ​$1600​.

Using continuous compounding, 6.71 years.

Step-by-step explanation:

Compound interest:

The compound interest formula is given by:

$A(t) = P(1 + \frac{r}{n})^{nt}$

Where A(t) is the amount of money after t years, P is the principal(the initial sum of money), r is the interest rate(as a decimal value), n is the number of times that interest is compounded per unit year and t is the time in years for which the money is invested or borrowed.

Continuous compounding:

The amount of money earned after t years in continuous interest is given by:

$P(t) = P(0)e^{rt}$

In which P(0) is the initial investment and r is the interest rate, as a decimal.

If Tanisha has ​$1000 to invest at 7​% per annum compounded semiannually​, how long will it be before she has ​$1600​?

We have to find t for which $A(t) = 1600$ when $P = 1000, r = 0.07, n = 2$

$A(t) = P(1 + \frac{r}{n})^{nt}$

$1600 = 1000(1 + \frac{0.07}{2})^{2t}$

$(1.035)^{2t} = \frac{1600}{1000}$

$(1.035)^{2t} = 1.6$

$\log{1.035)^{2t}} = \log{1.6}$

$2t\log{1.035} = \log{1.6}$

$t = \frac{\log{1.6}}{2\log{1.035}}$

$t = 6.83$

Using continuous interest 6.83 years before she has ​$1600​

If the compounding is​ continuous, how long will it​ be?

We have that $P(0) = 1000, r = 0.07$

Then

$P(t) = P(0)e^{rt}$

$1600 = 1000e^{0.07t}$

$e^{0.07t} = 1.6$

$\ln{e^{0.07t}} = \ln{1.6}$

$0.07t = \ln{1.6}$

$t = \frac{\ln{1.6}}{0.07}$

$t = 6.71$

Using continuous compounding, 6.71 years.