If $3000 is deposited at the end of each half year in an account that earns 6.2% compounded semiannually, how long will it be before the account contains $130,000?
To find out how long it will take for the account to contain $130,000, we can use the formula for the future value of an ordinary annuity.
The formula for the future value of an ordinary annuity is:
\(FV = P \times \dfrac{(1+r)^n - 1}{r}\)
where:
FV = future value (desired amount)
P = periodic payment
r = interest rate per period
n = number of periods
In this case, the periodic payment is $3000, the interest rate per period is 6.2% compounded semiannually (which means the interest rate per period is 6.2% / 2 = 3.1%), and we want to find the number of periods.
Let's substitute the given values into the formula:
\(130,000 = 3000 \times \dfrac{(1 + 0.031)^n - 1}{0.031}\)
Simplifying further:
\(43.33 = \dfrac{(1.031)^n - 1}{0.031}\)
Now, we can solve for \(n\) by isolating it.
\(43.33 \times 0.031 = (1.031)^n - 1\)
\(1.34123 = (1.031)^n - 1\)
Adding 1 to both sides:
\(2.34123 = (1.031)^n\)
To solve for \(n\), we take the logarithm of both sides:
\(log(2.34123) = log((1.031)^n)\)
Using the logarithm properties (logarithms with base 10):
\(log(2.34123) = n \times log(1.031)\)
Now, solve for \(n\) using a calculator:
\(n = \dfrac{log(2.34123)}{log(1.031)}\)
By plugging in the values in the calculator, we get:
\(n \approx 41.476\)
Therefore, it will take approximately 41.476 periods (or 20.738 years) for the account to reach $130,000.
To find out how long it will take for the account to reach $130,000, we need to use the formula for compound interest:
A = P(1 + r/n)^(nt),
Where:
A = Final amount ($130,000 in this case)
P = Principal amount ($3000 deposited at the end of each half-year)
r = Annual interest rate (6.2%, or 0.062 in decimal form)
n = Number of compounding periods per year (semiannually in this case, so n = 2)
t = Time (unknown variable we are trying to find)
Let's plug in these values and solve for t:
$130,000 = $3000(1 + 0.062/2)^(2t).
Divide both sides of the equation by $3000:
43.33 = (1 + 0.062/2)^(2t).
To isolate the exponent, take the logarithm of both sides. You may use either the common logarithm (log base 10) or the natural logarithm (log base e, denoted as ln).
For simplicity, let's use the natural logarithm:
ln(43.33) = ln[(1 + 0.031)^2t].
Using the property of logarithms, we can bring down the exponent 2t:
ln(43.33) = 2t ln(1 + 0.031).
Divide both sides of the equation by 2 ln(1 + 0.031):
t = ln(43.33) / (2 ln(1 + 0.031)).
Using a calculator, we can evaluate the right-hand side of the equation:
t ≈ 13.55.
Therefore, it will take approximately 13.55 years for the account to reach $130,000.
and what is your thinking?
3000(1.031^n - 1)/.031 = 130000
1.031^n - 1 = 1.3743333
1.031^n = 2.3743333
n log 1.031 = log 2.3743333
n = 28.32 half-years or appr. 14 years