A person invests 8000 dollars in a bank. The bank pays 6.25% interest compounded daily. To the nearest tenth of a year, how long must the person leave the money in the bank until it reaches 10800 dollars?
Let x be the number of years the money must be left in the bank.
Using the formula for compound interest, we have:
A = P(1 + r/n)^(nt)
Where:
A = the future value of the investment ($10800)
P = the principal amount ($8000)
r = the interest rate per period (6.25% or 0.0625)
n = number of times the interest is compounded per year (daily compounding means n = 365)
t = time in years (unknown)
Substitute the given values into the formula:
10800 = 8000(1 + 0.0625/365)^(365x)
Now, solve for x using algebra:
1.35 = (1 + 0.000171232)^(365x)
1.35 = (1.000171232)^(365x)
Take the natural logarithm of both sides:
ln(1.35) = ln(1.000171232)^(365x)
ln(1.35) = 365x * ln(1.000171232)
x = ln(1.35) / (365 * ln(1.000171232)
x ≈ 5.1 years
Therefore, the person must leave the money in the bank for approximately 5.1 years until it reaches $10800.