Chapter 3.5, Problem 22E

To determine

Tocomplete:

The table for time t, such that the investment P triples when the interest is compounded annually at rate r. Create a scatter plot.

Answer to Problem 22E

The completed table is:

  $\begin{array}{ccccccc}r& 2\%& 4\%& 6\%& 8\%& 10\%& 12\%\\ t& 55.48& 28.01& 18.85& 14.27& 11.53& 9.69\end{array}$

The scatter plot for this table is shown below:

  

Explanation of Solution

Given:

An incomplete table:

  $\begin{array}{ccccccc}r& 2\%& 4\%& 6\%& 8\%& 10\%& 12\%\\ t& & & & & & \end{array}$

Formula used:

Amount of anannually compounded account is given by:

  $A=P{\left(1+r\right)}^t$

Calculation:

We can set up the times for the different interest rates given in the table as shown below:

  $\begin{array}{l}r=0.02\Rightarrow 3P=P{\left(1+0.02\right)}^t\Rightarrow \ln \left(1.02\right)t=\ln 3\Rightarrow t=\frac{\ln 3}{\ln \left(1.02\right)}\approx 55.48\\ r=0.04\Rightarrow 3P=P{\left(1+0.04\right)}^t\Rightarrow \ln \left(1.04\right)t=\ln 3\Rightarrow t=\frac{\ln 3}{\ln \left(1.04\right)}\approx 28.01\\ r=0.06\Rightarrow 3P=P{\left(1+0.06\right)}^t\Rightarrow \ln \left(1.06\right)t=\ln 3\Rightarrow t=\frac{\ln 3}{\ln \left(1.06\right)}\approx 18.85\\ r=0.08\Rightarrow 3P=P{\left(1+0.08\right)}^t\Rightarrow \ln \left(1.08\right)t=\ln 3\Rightarrow t=\frac{\ln 3}{\ln \left(1.08\right)}\approx 14.27\\ r=0.10\Rightarrow 3P=P{\left(1+0.10\right)}^t\Rightarrow \ln \left(1.1\right)t=\ln 3\Rightarrow t=\frac{\ln 3}{\ln \left(1.1\right)}\approx 11.53\\ r=0.12\Rightarrow 3P=P{\left(1+0.12\right)}^t\Rightarrow \ln \left(1.12\right)t=\ln 3\Rightarrow t=\frac{\ln 3}{\ln \left(1.12\right)}\approx 9.69\end{array}$

Therefore, we can write the completed table as shown below:

  $\begin{array}{ccccccc}r& 2\%& 4\%& 6\%& 8\%& 10\%& 12\%\\ t& 55.48& 28.01& 18.85& 14.27& 11.53& 9.69\end{array}$

The scatter plot for this table is shown below: