Question

How do you calculate to the nearest hundredth of a year how long it takes for an amount of money to double if interest is compounded continuously at $ 3.8\% $ ?

Answer 1

Hint: In order to find the year for an amount of money to double if the interest is compounded continuously at some rate, use the formula $ A = P{e^{rt}} $ , where $ P $ , is the principal, $ r $ is the annual interest rate, $ t $ is the time in years, and $ A $ is the total amount at the end of the year. Just put the value in the formula and calculate accordingly.

Complete step by step solution:
We are given a rate of interest $ 3.8\% $ annually for an amount, which can be written in decimal as: $ 3.8\% = \dfrac{{3.8}}{{100}} = 0.038 $ .
Since, we know that the amount of some money with continuously compounded interest is given by the formula $ A = P{e^{rt}} $ , where $ P $ , is the principal, $ r $ is the annual interest rate, $ t $ is the time in years, and $ A $ is the total amount at the end of the year.
As, we are given double the amount, so the value of $ {e^{rt}} = 2 $ .
Substitute the value of $ r = 0.038 $ in the above equation and we get:
 $ {e^{0.038t}} = 2 $
Since, we need to find the time in years so, solving the above equation for $ t $ , for that taking log both sides and we get:
 $
  {e^{0.038t}} = 2 \\
  \ln \left( {{e^{0.038t}}} \right) = \ln 2 \;
  $
Further solving the equations:
 $
  \ln \left( {{e^{0.038t}}} \right) = \ln 2 \\
  0.038t\ln e = \ln 2 \;
  $
We know that the value of $ \ln 2 = 0.69314 $ and $ \ln e = 1 $ . Substituting these values above, solving it and we get:
 $
  0.038t\ln e = \ln 2 \\
  0.038t = 0.69314 \;
  $
Dividing both sides by to get the value of $ t $ :
 $
  \dfrac{{0.038t}}{{0.038}} = \dfrac{{0.69314}}{{0.038}} \\
  t = 18.240526 \;
  $
Since, we need to find the value up to nearest hundredth place so check out the hundred place after decimal and we get $ 4 $ in that place, so checking out the next right term after that and it is $ 0 $ which is less than $ 5 $ , so the value of $ 4 $ , remains same and rest all of terms on the right side of $ 4 $ becomes zero.
Therefore, the value of $ t = 18.24 $ .
Hence, the time it takes for an amount of money to double if interest is compounded continuously at $ 3.8\% $ is $ 18.24 $ years in the nearest hundredth place.
So, the correct answer is “ $ 18.24 $ years”.

Note: In place of time, if it was needed to find some other value, then also this formula would be used and the same steps would be followed.
We can also find the nearest tenth, thousand and other places also in a similar way.
Cross- check the answers once.
We use the following formula to find the required,
Formula used:
i. $ A = P{e^{rt}} $
ii. $ \log {x^y} = x\log y $
iii. $ \ln e = 1 $
iv. $ x\% = \dfrac{x}{{100}} $