Question

How many years would it take your money to double:(a) At 10% interest compounded ____yearly.(b) At 10% interest compounded weekly.____years and ___weeks(c) At 10% interest compounded continuously._____years.

Answer 1

The Solution:

Given that an amount of money is compounded yearly at 10% interest.

We are required to find how many years it will take the money to double.

By the Compound Interest formula,

$A=P(1+\frac{r}{100})^n$

Part a:

In this case,

$\begin{gathered} P=\text{ principal =?} \\ r=\text{ rate =10\%} \\ A=\text{ amount =2P} \\ n=\text{ number of years =?} \end{gathered}$

Substituting these values in the formula, we get

$2P=P(1+\frac{10}{100})^n$$\begin{gathered} 2P=P(1+0.1)^n \\ 2P=P(1.1)^n \end{gathered}$

Dividing through by P, we get

$2=(1.1)^n$

Taking the logarithm of both sides, we get

$\begin{gathered} \log _{}2=\log _{}(1.1)^n \\ \log _{}2=n\log _{}1.1 \end{gathered}$

Dividing both sides by log1.1, we get

$\begin{gathered} n=\frac{\log _{}2}{\log _{}1.1}=7.27254\approx7.3\text{ years} \\ \end{gathered}$

Therefore, it will take approximately 7.3 years to double your money.

Part b:

Compounded weekly.

$2P=P(1+\frac{r}{100\alpha})^{n\alpha}$

In this case,

$\begin{gathered} \alpha=\text{ number of week in year=52 week} \\ n=\text{?} \\ r=10\text{\%} \end{gathered}$

Substituting, we get

$2P=P(1+\frac{10}{100\times52})^{52n}$$2=(1+\frac{1}{520})^{52n}$$\begin{gathered} 2=(1.001923)^{52n} \\ \log 2=\log (1.001923)^{52n} \\ \frac{\log2}{\log(1.001923)}=52n \end{gathered}$$52n=360.7974145$

Dividing both sides, we get

$n=\frac{360.7974145}{52}=6.938412\approx6\text{ years and 49 weeks}$

Thus, it will take approximately 6 years and 49 weeks to double your money.

Part c:

Compounded continuously, which I suppose means compounded daily.

Recall:

365 days = 1 year

$\begin{gathered} 2P=P(1+\frac{r}{100\alpha})^{n\alpha} \\ \text{where} \\ \alpha=365\text{ days} \\ r=10\text{\%} \end{gathered}$

Substituting these values, we get

$\begin{gathered} 2P=P(1+\frac{10}{36500})^{365n} \\ \\ 2=(1+\frac{1}{3650})^{365n} \end{gathered}$$\begin{gathered} 2=(1.000273973)^{365n} \\ \log 2=\log (1.000273973)^{365n} \\ \log 2=365n\log (1.000273973) \end{gathered}$$\begin{gathered} 365n=\frac{\log 2}{\log (1.000273973)} \\ \\ 365n=2530.330098 \end{gathered}$

Dividing both sides by 365, we get

$n=\frac{2530.330098}{365}=6.932411\approx7\text{ years}$

Therefore, it will take approximately 7 years to double your money.