suppose you invest $500 in a savings account that pays 3.5% at an annual interest. When will the account contain $650?
I = PRT
150 = 500 * 0.035 * t
150 = 17.5t
150/17.5 = t
8.57 years = t
To find out when the account will contain $650, we need to determine the time period required to grow the initial investment of $500 to $650 at an annual interest rate of 3.5%.
To solve this, we can use the formula for compound interest:
A = P(1 + r/n)^(n*t)
Where:
A = the final amount in the account ($650)
P = the principal amount (initial investment of $500)
r = the annual interest rate (3.5% or 0.035 expressed as a decimal)
n = the number of times interest is compounded per year (assume yearly compounding)
t = the time in years
Let's plug in the values and solve for t:
$650 = $500(1 + 0.035/1)^(1*t)
Simplifying further:
1.3 = 1.035^t
To solve for t, we can take the logarithm of both sides using the natural logarithm (ln):
ln(1.3) = ln(1.035^t)
Using the logarithm properties, we can bring down the exponent:
ln(1.3) = t * ln(1.035)
Now we can isolate t by dividing both sides by ln(1.035):
t = ln(1.3) / ln(1.035)
Using a calculator, we find that t ≈ 4.44 (rounded to two decimal places).
Therefore, it will take approximately 4.44 years for the account to contain $650 with an annual interest rate of 3.5%.