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Sagot :
The annually compounded interest formula is:
[tex]A=P(1+\frac{r}{n})^{n\cdot t}[/tex]Where A is the amount you will have, P is the principal, r is the annual interest rate expressed as a decimal, n is the number of times the interest is compounded per time period (in this case as it is compounded annually, n=1) and t is the amount of time (in years) that the money is saved.
Then, we know that:
A=$8000
P=$4200
r=6.75%/100%=0.0675
n=1
t=?
Replace the known values and then solve for t as follows:
[tex]\begin{gathered} 8000=4200(1+0.0675)^t \\ 8000=4200(1.0675)^t \\ \frac{8000}{4200}=\frac{4200(1.0675)}{4200}^t \\ 1.9048=1.0675^t \\ Apply\text{ log to both sides} \\ \log 1.9048=\log 1.0675^t \\ By\text{ the properties of logarithms we have} \\ \log 1.9048=t\cdot\log 1.0675^{} \\ t=\frac{\log 1.9048}{\log 1.0675} \\ t=9.8647\approx10 \end{gathered}[/tex]Therefore, it will take 10 years to accumulate $8000 or more in the account.
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