If $4000 is deposited at the end of each year in an account that earns 6.2% compounded semiannually, how long will it be before the account contains $120,000?

Question

If $4000 is deposited at the end of each year in an account that earns 6.2% compounded semiannually, how long will it be before the account contains $120,000?

Answer 1

when is 4000(1.031)^n = 12000 ?

1.031^n = 3
take log of both sides, and use log rules

n log 1.031 = log 3
n = log3/log1.031 = appr 35.99
So it would take 36 half-years or 18 years

check:
4000(1.031)^36 = 12005.27 , not bad

Answer 2

To solve this problem, we can use the future value formula for compound interest:

\[FV = P(1 + \frac{r}{n})^{nt}\]

Where:
FV = future value
P = principal (initial deposit)
r = annual interest rate (in decimal form)
n = number of times interest is compounded per year
t = number of years

In this case, we have:
P = $4000
r = 6.2% = 0.062
n = 2 (since interest is compounded semiannually)

Let's substitute these values into the formula and solve for t:

$120,000 = $4000(1 + \frac{0.062}{2})^{2t}

Now, let's solve for t step by step:

1. Divide both sides of the equation by $4000:
$\frac{120000}{4000} = (1 + \frac{0.062}{2})^{2t}$

2. Simplify the left side:
30 = (1 + 0.031)^{2t}

3. Take the logarithm (base 10 or natural logarithm) of both sides to remove the exponent:
log(30) = log[(1 + 0.031)^{2t}]

4. Use the property of logarithms to bring down the exponent:
log(30) = 2t * log(1 + 0.031)

5. Divide both sides by log(1 + 0.031):
t = log(30) / (2 * log(1 + 0.031))

Now, we can evaluate this expression using a calculator to find the value of t.

Answer 3

To find out how long it will take for the account to contain $120,000, we need to use the formula for compound interest:

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

Where:
A = the future value of the investment (in this case, $120,000)
P = the principal amount (the initial deposit, in this case, $4000)
r = the annual interest rate (6.2%)
n = the number of times interest is compounded per year (semiannually means 2 times per year)
t = the number of years

Now let's solve for t:

120,000 = 4000(1 + 0.062/2)^(2t)

Divide both sides of the equation by 4000:

(1 + 0.062/2)^(2t) = 30

Take the logarithm of both sides of the equation:

log[(1 + 0.062/2)^(2t)] = log(30)

Using the logarithm property, we can bring the t down:

2t * log(1 + 0.062/2) = log(30)

Now we can solve for t by isolating it:

t = log(30)/[2 * log(1 + 0.062/2)]

Using a calculator, we can calculate t to be approximately 13.65 years.

Therefore, it will take approximately 13.65 years before the account contains $120,000.