A person invests 1000 dollars in a bank. The bank pays 6.75% interest compounded monthly. To the nearest tenth of a year, how long must the person leave the money in the bank until it reaches 2500 dollars?A, equals, P, left bracket, 1, plus, start fraction, r, divided by, n, end fraction, right bracket, start superscript, n, t, end superscriptA=P(1+ nr ) nt
Question
A person invests 1000 dollars in a bank. The bank pays 6.75% interest compounded monthly. To the nearest tenth of a year, how long must the person leave the money in the bank until it reaches 2500 dollars?A, equals, P, left bracket, 1, plus, start fraction, r, divided by, n, end fraction, right bracket, start superscript, n, t, end superscriptA=P(1+ nr ) nt
Solution
The formula you provided is a bit incorrect. The correct formula for compound interest 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 (in decimal).
- n is the number of times that interest is compounded per year.
- t is the time the money is invested for in years.
In this case, we know that:
- A = $2500
- P = $1000
- r = 6.75% = 0.0675 (in decimal)
- n = 12 (since the interest is compounded monthly)
We need to find t. So we will substitute the known values into the formula and solve for t.
1000(1 + 0.0675/12)^(12t)
Divide both sides by 1000:
2.5 = (1 + 0.0675/12)^(12t)
Take the natural logarithm (ln) of both sides:
ln(2.5) = ln((1 + 0.0675/12)^(12t))
Use the power rule of logarithms to bring down the exponent:
ln(2.5) = 12t * ln(1 + 0.0675/12)
Solve for t:
t = ln(2.5) / [12 * ln(1 + 0.0675/12)]
When you calculate the above expression, you will get the time in years to the nearest tenth.
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