A person invests 3500 dollars in a bank. The bank pays 4.5% interest compounded monthly. To the nearest tenth of a year, how long must the person leave the money in the bank until it reaches 6900 dollars?

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Answer:

$2.38

Step-by-step explanation:

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Time period for which person leave the money in the bank until it reaches the amount 6900 dollars is equals to 14.5 years.

What is compound interest?

"Compound interest is defined as the interest which we get on the accumulated amount of initial principal and interest which we have received for previous time duration."

Formula used

For Compound interest

[tex]A= P( 1 + \frac{r}{n}) ^{nt}[/tex]

A = final  amount

P = Principal

r = rate of interest

n= number of times interest applied per period time

t=number of time period elapsed

According to the question,

Given,

Principal 'P'= 3500 dollars

Rate of interest 'r' = 4.5%

Compounded monthly 'n' = 12

Final amount 'A' = 6900 dollars

Substitute the value in the formula of compound interest  to get the time,

[tex]6900 = 3500( 1+ \frac{4.5}{(12)(100))} )^{12t}[/tex]

⇒[tex]\frac{6900 }{ 3500} =( 1+ \frac{4.5}{(12)(100))} )^{12t}[/tex]

⇒[tex]\frac{69 }{ 35} =( \frac{803}{800})^{12t}[/tex]

⇒[tex]log69 -log35 = 12t( log 803-log800)[/tex]

⇒[tex]1.8388-1.5440=12t( 2.9047-2.903)[/tex]

⇒ [tex]0.2948 = 12t ( 0.0017)[/tex]

⇒[tex]t= \frac{2948}{(17)(12)}[/tex]

⇒[tex]t= 14.45[/tex]

⇒[tex]t=14.5[/tex] years

Hence, time period for which person leave the money in the bank until it reaches the amount 6900 dollars is equals to 14.5 years.

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