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+(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+(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+(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+(10)/(100*52))^(52n)`
`2=(1+(1)/(520))^(52n)`
`\begin{gathered} 2=(1.001923)^(52n) \\ \log 2=\log (1.001923)^(52n) \\ (\log2)/(\log(1.001923))=52n \end{gathered}`
`52n=360.7974145`

Dividing both sides, we get

`n=(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+(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+(10)/(36500))^(365n) \\ \\ 2=(1+(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=(\log 2)/(\log (1.000273973)) \\ \\ 365n=2530.330098 \end{gathered}`

Dividing both sides by 365, we get

`n=(2530.330098)/(365)=6.932411\approx7\text{ years}`

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