Question

Find how many years it would take for an investment of $4500 to grow to $7900 at an annual interest rate of 4.7% compounded daily.

Answer 1

To answer this question, we need to use the next formula for compound interest:

`A=P(1+(r)/(n))^(nt)`

From the formula, we have:

• A is the accrued amount. In this case, A = $7900.

,

• P is the principal amount. In this case, $4500.

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• r is the interest rate. In this case, we have 4.7%. We know that this is equivalent to 4.7/100.

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• n is the number of times per year compounded. In this case, we have that n = 365, since the amount is compounded daily.

Now, we can substitute each of the corresponding values into the formula as follows:

`A=P(1+(r)/(n))^(nt)\Rightarrow7900=4500(1+((4.7)/(100))/(365))^(365t)`

And we need to solve for t to find the number of years, as follows:

1. Divide both sides by 4500:

`(7900)/(4500)=(1+(0.047)/(365))^(365t)`

2. Applying natural logarithms to both sides (we can also apply common logarithms):

`\ln (7900)/(4500)=\ln (1+(0.047)/(365))^(365t)\Rightarrow\ln (7900)/(4500)=365t\ln (1+(0.047)/(365))`

3. Then, we have:

`(\ln(7900)/(4500))/(\ln(1+(0.047)/(365)))=365t\Rightarrow4370.84856503=365t`

4. And now, we have to divide both sides by 365:

`(4370.84856503)/(365)=t\Rightarrow t=11.9749275754`

If we round the answer to two decimals, we have that t is equal to 11.97 years.