Question

A person invests 9500 dollars in a bank. The bank pays 5.75% interest compounded annually. To the nearest tenth of a year, how long must the person leave the money in the bank until it reaches 22600 dollars?

Answer 1

10.6

Explanation:

To find out how long the person must leave the money in the bank until it reaches $22,600, we can use the formula for compound interest:

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

Where:

A = the final amount (22600 dollars in this case)

P = the principal amount (9500 dollars)

r = the annual interest rate (5.75% expressed as 0.0575)

n = the number of times the interest is compounded per year (in this case, annually, so n = 1)

t = the time in years

Substituting the given values into the formula, we have:

22600 = 9500(1 + 0.0575/1)^(1*t)

Simplifying:

2.3789 = (1.0575)^t

To solve for t, we take the logarithm of both sides:

log(2.3789) = log((1.0575)^t)

Using logarithm properties, we can bring the exponent down:

log(2.3789) = t * log(1.0575)

Now, we can solve for t by dividing both sides of the equation:

t = log(2.3789) / log(1.0575)

Using a calculator, we find:

t ≈ 10.6

Therefore, the person must leave the money in the bank for approximately 10.6 years to reach $22,600.