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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.

14. We are starting with $5000 and we’re going to put it into an account that earns an annual interest rate of 12%. How long should we leave the money in the account in order to double our money if interest is compounded

  1. quarterly
  2. monthly
  3. continuously
Identify the given quantities, plug into the appropriate equation and use the techniques from earlier problem to solve for \(t\).

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a quarterly Show Solution

From the problem statement we can see that,

\[A = 10000\hspace{0.5in}P = 5000\hspace{0.5in}\,\,r = \frac{{12}}{{100}} = 0.12\]

Remember that the value of \(r\) must be given as a decimal, i.e. the percentage divided by 100. Also, 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,

\[m = 4\]

Plugging into the equation gives us,

\[10000 = 5000{\left( {1 + \frac{{0.12}}{4}} \right)^{4t}} = 5000{\left( {1.03} \right)^{4t}}\]

Using the techniques from this section we can solve for \(t\).

\[\begin{align*}2 & = {1.03^{4t}}\\ \ln \left( 2 \right) & = \ln \left( {{{1.03}^{4t}}} \right)\\ \ln \left( 2 \right) & = 4t\ln \left( {1.03} \right)\\ t & = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{\ln \left( 2 \right)}}{{4\ln \left( {1.03} \right)}} = 5.8624}}\end{align*}\]

So, we’ll double our money in approximately 5.8624 years.


b monthly Show Solution

From the problem statement we can see that,

\[A = 10000\hspace{0.5in}P = 5000\hspace{0.5in}r = \frac{{12}}{{100}} = 0.12\]

Remember that the value of \(r\) must be given as a decimal, i.e. the percentage divided by 100. Also, for this part we are compounding interest rate monthly and that means it will compound 12 times per year so we also then know that,

\[m = 12\]

Plugging into the equation gives us,

\[10000 = 5000{\left( {1 + \frac{{0.12}}{{12}}} \right)^{12t}} = 5000{\left( {1.01} \right)^{12t}}\]

Using the techniques from this section we can solve for \(t\).

\[\begin{align*}2 & = {1.01^{12t}}\\ \ln \left( 2 \right) & = \ln \left( {{{1.01}^{12t}}} \right)\\ \ln \left( 2 \right) & = 12t\ln \left( {1.01} \right)\\ t & = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{\ln \left( 2 \right)}}{{12\ln \left( {1.01} \right)}} = 5.8051}}\end{align*}\]

So, we’ll double our money in approximately 5.8051 years.


c continuously Show Solution

From the problem statement we can see that,

\[A = 10000\hspace{0.5in}P = 5000\hspace{0.5in}\,\,r = \frac{{12}}{{100}} = 0.12\]

Remember that the value of \(r\) must be given as a decimal, i.e. the percentage divided by 100. For this part we are compounding continuously so we won’t have an \(m\) and will be using the other equation.

Plugging into the continuously compounding interest equation gives,

\[10000 = 5000{{\bf{e}}^{0.12t}}\]

Now, solving this gives,

\[\begin{align*}2 & = {{\bf{e}}^{0.12t}}\\ \ln \left( 2 \right) & = \ln \left( {{{\bf{e}}^{0.12t}}} \right)\\ \ln \left( 2 \right) & = 0.12t\\ t & = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{\ln \left( 2 \right)}}{{0.12}} = 5.7762}}\end{align*}\]

So, we’ll double our money in approximately 5.7762 years.