Question

$1750 is invested in an account earning 3.5% interest compounded annualy. How long will it need to be in an account to double?

Answer 1

Given :

`\begin{gathered} P\text{ = \$ 1750} \\ R\text{ = 3.5 \%} \\ A\text{ = 2P} \\ A\text{ = 2}*\text{ 1750 = \$ 3500} \end{gathered}`

Amount is given as,

`\begin{gathered} A\text{ = P( 1 + }(R)/(100))^T \\ 3500\text{ = 1750( 1 + }(3.5)/(100))^T \\ \text{( 1 + }(3.5)/(100))^T\text{ = }(3500)/(1720) \end{gathered}`

Further,

`\begin{gathered} \text{( 1 + }(3.5)/(100))^T\text{ = 2} \\ ((103.5)/(100))^T\text{ = }2 \\ (1.035)^T\text{ = 2} \end{gathered}`

Taking log on both the sides,

`\begin{gathered} \log (1.035)^T\text{ = log 2} \\ T\log (1.035)\text{ = log 2} \\ T\text{ = }\frac{\log \text{ 2}}{\log \text{ 1.035}} \end{gathered}`

Therefore,

`\begin{gathered} T\text{ = }(0.3010)/(0.0149) \\ T\text{ = 20.20 years }\approx\text{ 20 years} \end{gathered}`

Thus the required time is 20 years.