2.6: How long Does That Take?

Last updated
Save as PDF

Page ID: 148563

$ \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } $

$ \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} $

$ \newcommand{\dsum}{\displaystyle\sum\limits} $

$ \newcommand{\dint}{\displaystyle\int\limits} $

$ \newcommand{\dlim}{\displaystyle\lim\limits} $

$ \newcommand{\id}{\mathrm{id}}$ $ \newcommand{\Span}{\mathrm{span}}$

( \newcommand{\kernel}{\mathrm{null}\,}\) $ \newcommand{\range}{\mathrm{range}\,}$

$ \newcommand{\RealPart}{\mathrm{Re}}$ $ \newcommand{\ImaginaryPart}{\mathrm{Im}}$

$ \newcommand{\Argument}{\mathrm{Arg}}$ $ \newcommand{\norm}[1]{\| #1 \|}$

$ \newcommand{\inner}[2]{\langle #1, #2 \rangle}$

$ \newcommand{\Span}{\mathrm{span}}$

$ \newcommand{\id}{\mathrm{id}}$

$ \newcommand{\Span}{\mathrm{span}}$

$ \newcommand{\kernel}{\mathrm{null}\,}$

$ \newcommand{\range}{\mathrm{range}\,}$

$ \newcommand{\RealPart}{\mathrm{Re}}$

$ \newcommand{\ImaginaryPart}{\mathrm{Im}}$

$ \newcommand{\Argument}{\mathrm{Arg}}$

$ \newcommand{\norm}[1]{\| #1 \|}$

$ \newcommand{\inner}[2]{\langle #1, #2 \rangle}$

$ \newcommand{\Span}{\mathrm{span}}$ $ \newcommand{\AA}{\unicode[.8,0]{x212B}}$

$ \newcommand{\vectorA}[1]{\vec{#1}} % arrow$

$ \newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow$

$ \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } $

$ \newcommand{\vectorC}[1]{\textbf{#1}} $

$ \newcommand{\vectorD}[1]{\overrightarrow{#1}} $

$ \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} $

$ \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} $

$ \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } $

$\newcommand{\longvect}{\overrightarrow}$

$ \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} $

$\newcommand{\avec}{\mathbf a}$ $\newcommand{\bvec}{\mathbf b}$ $\newcommand{\cvec}{\mathbf c}$ $\newcommand{\dvec}{\mathbf d}$ $\newcommand{\dtil}{\widetilde{\mathbf d}}$ $\newcommand{\evec}{\mathbf e}$ $\newcommand{\fvec}{\mathbf f}$ $\newcommand{\nvec}{\mathbf n}$ $\newcommand{\pvec}{\mathbf p}$ $\newcommand{\qvec}{\mathbf q}$ $\newcommand{\svec}{\mathbf s}$ $\newcommand{\tvec}{\mathbf t}$ $\newcommand{\uvec}{\mathbf u}$ $\newcommand{\vvec}{\mathbf v}$ $\newcommand{\wvec}{\mathbf w}$ $\newcommand{\xvec}{\mathbf x}$ $\newcommand{\yvec}{\mathbf y}$ $\newcommand{\zvec}{\mathbf z}$ $\newcommand{\rvec}{\mathbf r}$ $\newcommand{\mvec}{\mathbf m}$ $\newcommand{\zerovec}{\mathbf 0}$ $\newcommand{\onevec}{\mathbf 1}$ $\newcommand{\real}{\mathbb R}$ $\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}$ $\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}$ $\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}$ $\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}$ $\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}$ $\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}$ $\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}$ $\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}$ $\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}$ $\newcommand{\laspan}[1]{\text{Span}\{#1\}}$ $\newcommand{\bcal}{\cal B}$ $\newcommand{\ccal}{\cal C}$ $\newcommand{\scal}{\cal S}$ $\newcommand{\wcal}{\cal W}$ $\newcommand{\ecal}{\cal E}$ $\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}$ $\newcommand{\gray}[1]{\color{gray}{#1}}$ $\newcommand{\lgray}[1]{\color{lightgray}{#1}}$ $\newcommand{\rank}{\operatorname{rank}}$ $\newcommand{\row}{\text{Row}}$ $\newcommand{\col}{\text{Col}}$ $\renewcommand{\row}{\text{Row}}$ $\newcommand{\nul}{\text{Nul}}$ $\newcommand{\var}{\text{Var}}$ $\newcommand{\corr}{\text{corr}}$ $\newcommand{\len}[1]{\left|#1\right|}$ $\newcommand{\bbar}{\overline{\bvec}}$ $\newcommand{\bhat}{\widehat{\bvec}}$ $\newcommand{\bperp}{\bvec^\perp}$ $\newcommand{\xhat}{\widehat{\xvec}}$ $\newcommand{\vhat}{\widehat{\vvec}}$ $\newcommand{\uhat}{\widehat{\uvec}}$ $\newcommand{\what}{\widehat{\wvec}}$ $\newcommand{\Sighat}{\widehat{\Sigma}}$ $\newcommand{\lt}{<}$ $\newcommand{\gt}{>}$ $\newcommand{\amp}{&}$ $\definecolor{fillinmathshade}{gray}{0.9}$

INTRODUCTION

Let's spend a few minutes at the start of this collaboration to review some important skills and facts from Preparation M.6. Most importantly:

Solving equations algebraically is all about “undoing” various arithmetic operations. It could be as simple as dividing/multiplying both sides of the given equation, or adding/subtracting a number from both sides of the given equation.

However, as we study more advanced mathematical models, we’ll need more advanced operations. Specifically, square roots, cube roots, and logarithms.

If we want to undo the squaring operation, we need a special operation, the square root
If we want to undo the cubing operation, we need a special operation, the cube root
If we want to undo an operation when the variable is in the exponent, we need to use logarithms

SPECIFIC OBJECTIVES

By the end of this collaboration, you should understand that

the log operation undoes the “raise to a power” operation.
exponential equations can be solved algebraically using logarithms.

By the end of this collaboration, you should be able to

solve exponential equations using logarithms.
check solutions to exponential equations obtained algebraically.

PROBLEM SITUATION: SOLVING EXPONENTIAL EQUATIONS

We use “opposite” operations to solve equations. If you can see what operations are currently applied to a variable, you can do the opposite operations to solve the equation—to find out what number the variable stands for. Some examples of operations in math are: addition, subtraction, multiplication, division, raising a number to a power, and taking a root of a number (such as the square root).

(1) Fill in the column on the right in the table below:

If you have an equation like...	...you can solve it by using the opposite operation. Write the operation you would use to solve the equation, and then solve it.
$x+5.4 = 11.6$
$t - 28 = 44$
$5w = 38$
$\dfrac{d}{12} = 3.5$

(2) In this course, we have also learned that there is a way to “undo” exponents. Fill in the column on the right in the table below:

If you have an equation like...

...you can solve it by using the opposite operation.

Write the operation you would use to solve the equation, and then solve it.

$x^2 = 36$

$t^3 = 27$

Exponential equations

Now, consider the following example. Even though both equations include the same numbers and the same variable, they are fundamentally different. It’s all about the placement of a variable.

The equation on the left is a cubic equation; the variable x is in the base of x3
The equation on the right is an exponential equation; the variable x is in the exponent of 3x

Given equation	x³=25	3^x=25
Given operation	The variable x is cubed	3 is raised to the power of x
“Opposite” operation	Cube root	Logarithm

An equation that has a variable in the exponent is called an exponential equation. We will be discussing such equations in more detail throughout this collaboration.

Sometimes, we can solve such equations by inspection. This means we look at an equation and see what number could go in place of the variable to make the equation true, such as in the equation 10^x= 100, where we know that x = 2 because 10²= 100.

(3) Solve the equations below by inspection.

Solve for x: 5^x=25

Solve for x: 3^x=81

In Preparation M.6, we saw that we can also undo the operation in an exponential equation using logarithms, which are often referred to as logs. We learned to use the log operation to solve equations when the base is the number 10 (such as 10^x=16).

(4) Solve the equations below. Think about the operation you would use to solve the equation first and then solve it. Round to the nearest thousandth.

10^x=135

10^x=12.2

What happens if the base is something other than 10?

As we saw in Question 3 above, if you have the equation: 7^x= 49, you might be able to use inspection to see that x = 2 because you remember that 7²=49. But what about an equation like 1.05^x = 1.6?

This kind of equation cannot be solved by remembering multiplication facts that we’ve learned before, or by a simple use of the “Log” key on a calculator, since the calculator assumes a base of 10 (not 1.05). We need to have a way to undo the exponent in this type of equation.

To solve an exponential equation, we can use the following formula:

\[ If\;a^x = b,\;then\;x = \dfrac{log\;b}{log\;a} \nonumber \]

where a and b are numbers, a is the base of the exponential equation, and x is a variable.

If you have an equation like...	...you can solve it using this new formula.
$7^x = 49$	We use logarithms to solve this equation: $ x = \dfrac{log\;49}{log\;7}$ Use a calculator to check that you get the solution $x = 2$
If you have an equation like...	...you can solve it using this new formula.
$1.05^t = 1.6$	The way to use logarithms to solve this equation is: $t = \dfrac{log\;1.6}{log\;1.05}$ And when you put this into a calculator, you should get a solution like t ≈ 9.6331635...= 9.633 rounded to thousandths

(5) Use the new formula to solve the equations below. Round to the nearest thousandth.

3^x= 27

6^t=200

1.032^t=2

(6) Note that the first equation in Question 5 could have been solved by inspection. Without using the new formula, how could you find an estimate for the second equation in the table?

Summary of Exponential Equations and Logarithms

Let’s summarize what we know about exponential equations and logarithms so far:

Exponential equations usually fit the format a = b^x where a and b are constant.
For example, 10^x = 24.
Depending on the value of b, we solve exponential equations in two ways:

Base	$10$	$a$
Exponential equation	$b = 10^x$	$b = a^x$
Solution	$x = log\;b$	$x = \dfrac{log\;b}{log\;a}$
Example	$Solve\;2 = 10^x$ $x = log\;2$ $x = 0.30$	$Solve\;15 = 2^x$ $x = \dfrac{log\;15}{log\;2}$ $x = 3.91$

Later in this collaboration and in the corresponding exercise assignment, you will see exponential equations that don’t quite the format of a = b^x. When that happens, try to simplify it first using addition, subtraction, multiplication or division.

Applications of Exponential Equations and Logarithms

Recall the following scenario from Collaboration M.5:

Suppose that the cost of attending a certain four-year private college (tuition and fees) was $16,500 in 1977 and that the cost increased about 3.2% every year. In Collaboration M.5, we found the model for the situation. We then thought about how we might determine the year when tuition would reach $42,000. We considered various methods, such as guess-and-check (guessing different values and checking the results), to find a solution to the equation:

\[42,000 = 16,500(1.032)^t \nonumber \]

where t = 0 in 1977.

(7) In what year is tuition predicted to be $42,000?

(8) What method could you use to determine whether your solution is correct? Explain your method, then use the method to check your answer from Question 7.

Exponential equations are often seen when working with accounts or loans that use compound interest. Compound interest occurs when interest is calculated on both the initial principal and any accumulated interest from previous periods. (Note: For comparison, recall that simple interest does not take into account the accumulated interest, that is, the interest each year is only based on the initial principal.) The equation for computing the amount of a loan, or in an account, that uses compound interest (assuming no change in principal) is below:

\[A = P(1 + \frac{r}{n})^{nt} \nonumber \]

In this formula,

A represents the total amount of the loan or in the account.
P represents the principal (initial amount of the loan or money put into the account).
r represents the interest rate, written as a decimal.
n represents the number of compounding periods per year.
t represents the term of the loan in years or time since opening the account in years.

(9) Suppose that you take out a $2000 loan with a compound interest rate of 4%, compounded annually (once a year). If you make no payments, and interest compounds annually, in how many years will the loan balance double (reach or surpass $4000)? Think about this on your own for a minute before sharing your ideas in your group.

Hint: when n = 1, the formula simplifies to: A = P(1 + r)^t.

(10) Jason has just consumed a 16-ounce cup of coffee containing 300 mg of caffeine.

(a) If caffeine is being eliminated from Jason’s body at a rate of 13% every hour, how long before three quarters of the caffeine has been eliminated? Enter your answer rounded to the nearest hour. (Hint: First, set up a model for the situation and then use it to answer the question.)

(b) How much caffeine will be in Jason’s blood if he drinks another 16-ounce cup of coffee 5 hours after the first?

MAKING CONNECTIONS

Record the important mathematical ideas from the discussion.

If you have an equation like...	...you can solve it by using the opposite operation. Write the operation you would use to solve the equation, and then solve it.
\(x+5.4 = 11.6\)
\(t - 28 = 44\)
\(5w = 38\)
\(\dfrac{d}{12} = 3.5\)

Base	\(10\)	\(a\)
Exponential equation	\(b = 10^x\)	\(b = a^x\)
Solution	\(x = log\;b\)	\(x = \dfrac{log\;b}{log\;a}\)
Example	\(Solve\;2 = 10^x\) \(x = log\;2\) \(x = 0.30\)	\(Solve\;15 = 2^x\) \(x = \dfrac{log\;15}{log\;2}\) \(x = 3.91\)

Search

Text Color

Text Size

Margin Size

Font Type

If you have an equation like...	...you can solve it using this new formula.
\(7^x = 49\)	We use logarithms to solve this equation: \( x = \dfrac{log\;49}{log\;7}\) Use a calculator to check that you get the solution \(x = 2\)
If you have an equation like...	...you can solve it using this new formula.
\(1.05^t = 1.6\)	The way to use logarithms to solve this equation is: \(t = \dfrac{log\;1.6}{log\;1.05}\) And when you put this into a calculator, you should get a solution like t ≈ 9.6331635...= 9.633 rounded to thousandths