Determine when, to the nearest year, $3,000 invested at 5% per year, compounded daily, will be worth $10,000.
10000 = 3000(1 + .05/365)^(365x)
x = 24.08
or, 24 years
To determine when $3,000 invested at 5% per year, compounded daily, will be worth $10,000, we can use the formula for compound interest:
A = P(1 + r/n)^(nt)
Where:
A = the future value of the investment
P = the principal amount (initial investment)
r = the annual interest rate (in decimal form)
n = the number of times that interest is compounded per year
t = the number of years
In this case, the amount we want is $10,000 (A), the principal is $3,000 (P), the interest rate is 5% (0.05 in decimal form), and interest is compounded daily (n = 365). Let's find out the time it takes (t) to reach $10,000.
Using the formula, we rewrite it as:
10,000 = 3,000(1 + 0.05/365)^(365t)
Divide both sides of the equation by 3,000:
10,000 / 3,000 = (1 + 0.05/365)^(365t)
Simplify the left side:
10/3 = (1 + 0.05/365)^(365t)
Now, we need to isolate the exponent. To do this, we take the natural logarithm (ln) of both sides:
ln(10/3) = ln[(1 + 0.05/365)^(365t)]
Using the property of logarithms, we bring down the exponent:
ln(10/3) = (365t) ln(1 + 0.05/365)
Now, we solve for t by dividing both sides by 365 and then dividing by ln(1 + 0.05/365):
t = [ln(10/3)] / [365 ln(1 + 0.05/365)]
Using a calculator, we can substitute the values and solve for t:
t ≈ 15.477
Therefore, to the nearest year, it will take approximately 15 years for the $3,000 investment to grow to $10,000 when compounded daily at an interest rate of 5% per year.