Compound interest is calculated on both the initial payment and the interest earned in previous periods.
problem
Suppose that a savings account is compounded yearly with a principal of $3000000. After 2 years years, the amount increased to $3421872. What was the per annum interest rate?
solution
Interest rate per anum was 6.8%.
explanation
To find interest rate we use formula:
$$ A = P \left( 1 + \frac{r}{n} \right)^{\Large{n \cdot t}} $$ |
A = total amount P = principal (amount of money deposited) r = annual interest rate n = number of times compounded per year t = time in years |
In this example we have
$$ A = $3421872 ~,~ P = $3000000 , t = 2 ~ \text{years} ~~ \text{and} ~ n = 1$$After plugging the given information we have
$$ \begin{aligned} 3421872 &= 3000000 \left( 1 + \frac{ r }{ 1 } \right)^{\Large{ 1 \cdot 2 }} \\ 3421872 &= 3000000 \left( 1 + \frac{ r }{ 1 } \right)^{\Large{ 2 }} \\ \left( 1 + \frac{ r }{ 1 } \right)^{\Large{ 2 }} &= \frac{ 3421872 }{ 3000000 }\\ \left( 1 + \frac{ r }{ 1 } \right)^{\Large{ 2 }} &= 1.1406 ~~~ \text{ Take the natural logarithm of each side} \\ ln \left( 1 + \frac{ r }{ 1 }\right) ^{\Large{ 2 }} &= ln(1.1406) \\ 2 \cdot ln \left( 1 + \frac{ r }{ 1 }\right) &= ln(1.1406) \\ ln \left( 1 + \frac{ r }{ 1 }\right) &= \frac{ln(1.1406)}{ 2} \\ ln \left( 1 + \frac{ r }{ 1 }\right) &= 0.0658 \\ 1 + \frac{ r }{ 1 } &= e^{ 0.0658 } \\ 1 + \frac{ r }{ 1 } &= 1.068 \\ \frac{ r }{ 1 } &= 0.068 \\ r &= 0.068 \approx 6.8\% \end{aligned}$$Please tell me how can I make this better.