Question

A person invests 7500 dollars in a bank. The bank pays 6% interest compounded semi-annually. To the nearest tenth of a year, how long must the person leave the money in the bank until it reaches 16200 dollars?

Answer 1

To determine how long to leave $7500 in the bank at 6% interest compounded semi-annually to reach $16200, use the compound interest formula. Solving for t, time in years, yields approximately 11.9 years for the investment to grow to the desired amount.

Step-by-step explanation:

To calculate how long the person must leave their money in the bank for it to grow from $7500 to $16200 with an interest rate of 6% compounded semi-annually, we'll use the formula for compound interest:

A = P(1 + \frac{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 (decimal).
• n is the number of times that interest is compounded per year.
• t is the number of years the money is invested for.

We are given:

• A = $16200
• P = $7500
• r = 0.06 (since 6% must be converted to a decimal)
• n = 2 (because the interest is compounded semi-annually)

Our goal is to solve for t, the time in years. Plugging the known values into the formula:

$16200 = $7500(1 + \frac{0.06}{2})^{2t}

Divide both sides by $7500:

2.16 = (1 + \frac{0.06}{2})^{2t}

Now take the natural logarithm of both sides:

ln(2.16) = ln((1 + \frac{0.06}{2})^{2t})

Use properties of logarithms to bring down the exponent:

ln(2.16) = 2t \cdot ln(1 + \frac{0.06}{2})

Divide both sides by 2ln(1 + \frac{0.06}{2}) to solve for t:

t = \frac{ln(2.16)}{2 \cdot ln(1 + \frac{0.06}{2})}

t = \frac{ln(2.16)}{2 \cdot ln(1.03)}

Using a calculator, we find t ≈ 11.9 years. Therefore, the person must leave the money in the bank for approximately 11.9 years to reach $16200.

Answer 2

`13.0\ years`

Step-by-step explanation:

we know that

The compound interest formula is equal to

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

where

A is the Final Investment Value

P is the Principal amount of money to be invested

r is the rate of interest in decimal

t is Number of Time Periods

n is the number of times interest is compounded per year

in this problem we have

`t=?\ years\\ P=\$7,500\\ r=0.06\\n=2\\ A=\$16,200`

substitute in the formula above

`16,200=7,500(1+(0.06)/(2))^(2t)`

`(2.16)=(1.03)^(2t)`

Apply log both sides

`log(2.16)=(2t)log(1.03)`

`t=log(2.16)/[(2)log(1.03)]`

`t=13.0\ years`