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A sum of money is invested at $12\%$ compounded quarterly. About how long will it take for the amount of money to double?



Compound interest formula: $V(t)=P\left(1+\frac{r}{n}\right)^{n t}$

$t=$ years since initial deposit

$n=$ number of times compounded per year

$r=$ annual interest rate (as a decimal)

$P=$ initial (principal) investment

$V(t)=$ value of investment after $t$ years


A23.4 years
B5.9 years
C6.1 years
D24.5 years

Verified Answer :

  • Define the investment growth using the formula \(2P = P(1 + \frac{0.12}{4})^{4t}\).
  • Simplify to \(2 = (1.03)^{4t}\) and apply logarithms to isolate \(t\).
  • Solve the equation \(t = \frac{\ln(2)}{4 \cdot \ln(1.03)}\).
  • Calculate the result \(t \approx 5.86\), which rounds to \(\boxed{5.9}\) years.

Explanation

  1. Problem Analysis and Setup

To determine how long it will take for an investment to double with compound interest, we start with the compound interest formula:
\[V(t) = P\left(1 + \frac{r}{n}\right)^{nt}\]
Given the parameters:

  • \(V(t) = 2P\) (the amount doubles)
  • \(r = 0.12\) (the annual interest rate)
  • \(n = 4\) (the interest is compounded quarterly)
  • \(P\) is the principal investment.

Our goal is to solve for \(t\), the time in years required for the investment to grow to twice its initial value.

  1. Logarithmic Formulation

We begin by substituting the known values into the compound interest formula:
\[2P = P\left(1 + \frac{0.12}{4}\right)^{4t}\]
First, divide both sides by \(P\) (assuming \(P > 0\)):
\[2 = \left(1 + 0.03\right)^{4t}\]
\[2 = (1.03)^{4t}\]
To solve for \(t\) in the exponent, we take the natural logarithm (\(\ln\)) of both sides:
\[\ln(2) = \ln((1.03)^{4t})\]
Using the logarithmic identity \(\ln(a^b) = b \cdot \ln(a)\), we rewrite the equation:
\[\ln(2) = 4t \cdot \ln(1.03)\]

  1. Calculation and Conclusion

To isolate \(t\), we divide both sides by \(4 \cdot \ln(1.03)\):
\[t = \frac{\ln(2)}{4 \cdot \ln(1.03)}\]
Using the values \(\ln(2) \approx 0.6931\) and \(\ln(1.03) \approx 0.02956\), we calculate the value of \(t\):
\[t \approx \frac{0.6931}{4 \cdot 0.02956} \approx \frac{0.6931}{0.11824} \approx 5.8624\]
Rounding to the nearest tenth, we get \(t \approx 5.9\) years. Comparing this result with the provided options, we see that it aligns with option (b).

Examples

This calculation is highly relevant in personal finance, for instance, when deciding where to place your savings. If you are comparing two different bank accounts that offer interest compounded at different intervals, understanding the power of compounding allows you to estimate how quickly your money will grow. This knowledge helps you make informed decisions about your financial future, ensuring your savings work effectively for you over time.

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