Question

Rachel deposited $6,449.82 into a savings account with an interest of 6.9% compounded annually. About how long will it take for the accout to be worth $8,000?

Answer 1

Answer:

3 years and 3 months.

Step-by-step explanation:

Since, the amount formula in compound interest,

$A= P(1+\frac{r}{100})^t$

Where,

P = principal amount,

r = rate per period,

t = number of periods,

Here, A = $ 8,000, r = 6.9%, P = $6,449.82,

By substituting the values,

$8000 = 6449.82(1+\frac{6.9}{100})^t$

$\frac{8000}{6449.82}=(1+\frac{6.9}{100})^t$

$\ln (\frac{8000}{6449.82})=t\ln(1.069)$

$\implies t = \frac{\ln (\frac{8000}{6449.82})}{\ln(1.069)}=3.228$

Hence, it would be take 3.228 years or 3 years and 3 months.

Answer 2

You will need this formula:

Years = {log(total) -log(Principal)} ÷ log(1 + rate)

Years = {log (8,000.00) - log ( 6,449.82 )} ÷ log (1.069)

Years =  (3.903089987 - 3.8095475946) / 0.028977705209

Years =  0.0935423924 / 0.028977705209

Years =  3.2280814414

Which would be approximately 3 years 3 months

Source:
http://www.1728.org/compint2.htm