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Section 1-9 : Exponential And Logarithm Equations

Compound Interest. If we put \(P\) dollars into an account that earns interest at a rate of \(r\) (written as a decimal as opposed to the standard percent) for \(t\) years then,

  1. if interest is compounded \(m\) times per year we will have, \[A = P{\left( {1 + \frac{r}{m}} \right)^{t\,m}}\]

    dollars after \(t\) years.

  2. if interest is compounded continuously we will have, \[A = P{{\bf{e}}^{r\,t}}\]

    dollars after \(t\) years.

13. We have $10,000 to invest for 44 months. How much money will we have if we put the money into an account that has an annual interest rate of 5.5% and interest is compounded

  1. quarterly
  2. monthly
  3. continuously
Hint : There really isn’t a whole lot to these other than to identify the quantities and then plug into the appropriate equation and compute the amount. Also note that you’ll need to make sure that you don’t do too much in the way of rounding with the numbers here. A little rounding can lead to very large errors in these kinds of computations.

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a quarterly Show Solution

From the problem statement we can see that,

\[P = 10000\hspace{0.5in}r = \frac{{5.5}}{{100}} = 0.055\hspace{0.5in}t = \frac{{44}}{{12}} = \frac{{11}}{3}\]

Remember that the value of \(r\) must be given as a decimal, i.e. the percentage divided by 100. Also remember that \(t\) must be in years so we’ll need to convert to years.

For this part we are compounding interest rate quarterly and that means it will compound 4 times per year so we also then know that,

\[m = 4\]

At this point all that we need to do is plug into the equation and run the numbers through a calculator to compute the amount of money that we’ll have.

\[A = 10000{\left( {1 + \frac{{0.055}}{4}} \right)^{\frac{{11}}{3}\left( 4 \right)}} = 10000{\left( {1.01375} \right)^{\frac{{44}}{3}}} = 10000\left( {1.221760422} \right) = 12217.60\]

So, we’ll have $12,217.60 in the account after 44 months.


b monthly Show Solution

From the problem statement we can see that,

\[P = 10000\hspace{0.5in}r = \frac{{5.5}}{{100}} = 0.055\hspace{0.5in}t = \frac{{44}}{{12}} = \frac{{11}}{3}\]

Remember that the value of \(r\) must be given as a decimal, i.e. the percentage divided by 100. Also remember that \(t\) must be in years so we’ll need to convert to years.

For this part we are compounding interest rate monthly and that means it will compound 12 times per year and so we also then know that,

\[m = 12\]

At this point all that we need to do is plug into the equation and run the numbers through a calculator to compute the amount of money that we’ll have.

\[A = 10000{\left( {1 + \frac{{0.055}}{{12}}} \right)^{\frac{{11}}{3}\left( {12} \right)}} = 10000{\left( {1.00453333} \right)^{44}} = 10000\left( {1.222876562} \right) = 12228.77\]

So, we’ll have $12,228.77 in the account after 44 months.


c continuously Show Solution

From the problem statement we can see that,

\[P = 10000\hspace{0.5in}r = \frac{{5.5}}{{100}} = 0.055\hspace{0.5in}t = \frac{{44}}{{12}} = \frac{{11}}{3}\]

Remember that the value of \(r\) must be given as a decimal, i.e. the percentage divided by 100. Also remember that \(t\) must be in years so we’ll need to convert to years.

For this part we are compounding continuously so we won’t have an \(m\) and will be using the other equation and all we have all we need to do the computation so,

\[A = 10000{{\bf{e}}^{\left( {0.055} \right)\left( {\frac{{11}}{3}} \right)}} = 10000{{\bf{e}}^{0.2016666667}} = 10000\left( {1.223440127} \right) = 12234.40\]

So, we’ll have $12,234.40 in the account after 44 months.