Determine when, to the nearest year, $3,000 invested at 5% per year, compounded daily, will be worth $10,000.

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

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

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

We know that A = $10,000, P =$3,000, r = 5% or 0.05 (in decimal), and n = 365 (compounded daily). We want to find t.

So, we can set up the equation as follows:

$10,000 =$3,000 * (1 + 0.05/365)^(365*t)

First, divide both sides by $3,000:

10,000/3,000 = (1 + 0.05/365)^(365*t)

3.3333 = (1 + 0.05/365)^(365*t)

Next, take the natural logarithm (ln) of both sides:

ln(3.3333) = 365*t * ln(1 + 0.05/365)

Then, divide both sides by 365 * ln(1 + 0.05/365) to solve for t:

t = ln(3.3333) / (365 * ln(1 + 0.05/365))

When you calculate the right side, you get approximately 22.57 years.

So, to the nearest year, $3,000 invested at 5$10,000 in 23 years.