Jordan invested $8,500 in an account paying an interest rate of 4.1% compounded daily. Assuming no deposits or withdrawals are made, how long would it take, to the nearest year, for the value of the account to reach $13,070?

 To solve this problem, we will use the formula for compound interest, which is:

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

Where: - A is the amount of money accumulated after n years, including interest. - P is the principal amount (the initial amount of money). - r is the annual interest rate (decimal). - n is the number of times that interest is compounded per year. - t is the time the money is invested for, in years.

We are given: - A = $13,070 (the amount of money accumulated we want to reach) - P = $8,500 (the initial principal balance) - r = 4.1% = 0.041 (the annual interest rate in decimal) - n = 365 (since the interest is compounded daily)

We want to find t (the time it will require for the account to grow to $13,070). We'll insert the values we have into the formula and solve for t:

$13,070 = $8,500(1 + 0.041/365)^(365t)

Divide both sides by $8,500 to isolate the exponential part:

1.53764706 = (1 + 0.041/365)^(365t)

To solve for t, we'll take the natural logarithm (ln) of both sides of the equation:

ln(1.53764706) = ln((1 + 0.041/365)^(365t))

The property of logarithms allows us to bring the exponent down:

ln(1.53764706) = 365t * ln(1 + 0.041/365)

Now, we will solve for t:

t = ln(1.53764706) / (365 * ln(1 + 0.041/365))

Use a calculator for the computation:

t ≈ ln(1.53764706) / (365 * ln(1.00011232876712)) t ≈ 0.4292196 / (365 * 0.00011232876712) t ≈ 0.4292196 / 0.0409999768 t ≈ 10.47233837

Rounded to the nearest year, it will take approximately 10 years for the value of the account to reach $13,070.