Section 6.3 Logarithms in Applications

Objectives

After completing this section, you should be able to do the following.

Convert an exponential function between standard form $Q = A \cdot b^t$ and base-$e$ form $Q = A \cdot e^{kt}\text{.}$
Use logarithms to calculate the doubling time and the half-life of an exponentially growing or decaying quantity.
Apply logarithms to solve application problems involving exponential growth and decay, including pH calculations.

Logarithms appear in many applied problems. They are indispensable in the context of exponential growth and decay.

Subsection Converting Between $Q = A \cdot b^t$ and $Q = A \cdot e^{kt}$

In Chapter 5 we considered exponential functions in two forms: the standard form $Q(t) = A \cdot b^t$ and the base-$e$ form $Q(t) = A \cdot e^{kt}\text{.}$ Recall that we called $k$ the “continuous growth rate”. In Section 5.5, it was shown that you can convert an exponential function given in base-$e$ form to standard form by simply taking $b = e^k\text{:}$

\begin{equation*} Q(t) = A \cdot e^{kt} = A \cdot (e^k)^t = A \cdot b^t. \end{equation*}

To convert an exponential function from standard form $Q(t) = A \cdot b^t$ to base-$e$ form, we have to find $k$ such that $e^k = b\text{.}$ If we have such a constant $k\text{,}$ we can easily rewrite $Q(t)$ in base-$e$ form:

\begin{equation*} Q(t) = A \cdot b^t = A \cdot (e^k)^t = A \cdot e^{kt}. \end{equation*}

Therefore, we need the solution $k$ to the equation $e^k = b\text{.}$ Applying the natural logarithm to both sides gives $\ln(e^k) = \ln(b)\text{,}$ and then Property 5 implies $k \ln(e) = \ln(b)\text{.}$ Since $\ln(e) = 1\text{,}$ the solution is

\begin{equation*} k = \ln(b). \end{equation*}

Result 6.3.1. Converting An Exponential Function To And From Base $e$.

Consider any arbitrary exponential function $Q(t) = A \cdot e^{kt}\text{.}$

Using $b = e^k\text{,}$ then $Q(t) = A \cdot b^t\text{.}$
Using $k = \ln b\text{,}$ then $Q(t) = A \cdot e^{kt}\text{.}$
If $b \gt 1\text{,}$ $\ln(b) \gt 0\text{,}$ so $k \gt 0\text{.}$
If $0 \lt b \lt 1\text{,}$ $\ln(b) \lt 0\text{,}$ so $k \lt 0\text{.}$

The last two bullet points follow from the properties of the natural logarithm function $y = \ln(x)\text{.}$ For inputs between $0$ and $1\text{,}$ $\ln(x)$ is negative. For inputs greater than $1\text{,}$ $\ln(x)$ is positive. Recall the graph of $y = \ln(x)\text{:}$

Example 6.3.2.

Rewrite the following exponential functions in terms of the natural base:

$\displaystyle P(t) = 200 \cdot 1.089^t$
$\displaystyle Q(t) = 1500 \cdot 0.72^t$

Solution.

With the result above, we can rewrite $P(t)$ in the form $P(t) = 200 e^{kt}$ where $k = \ln 1.089 \approx 0.085\text{.}$ Therefore,

\begin{equation*} P(t) = 200 e^{0.085t} \end{equation*}

and $k$ is positive as $b = 1.089 \gt 1\text{.}$
Letting $k = \ln 0.72 \approx -0.329\text{,}$ then

\begin{equation*} Q(t) = 1500 e^{-0.329t}. \end{equation*}

Notice that $k$ is negative since the growth factor $b = 0.72$ is less than $1\text{.}$

Note: Whenever you calculate $k$ or the growth factor $b\text{,}$ do not round off too crudely. Take at least three or four decimal places. Using too few decimal places to approximate $k$ or $b$ may lead to significant errors in the values of an exponential function.

Example 6.3.3.

The common antianxiety medication Diazepam decays in a patient’s body after a dose is taken. Suppose a patient takes a single dose of $10$ mg of Diazepam. The amount $D(t)$ left after $t$ days is given by:

\begin{equation*} D(t) = 10 (0.74)^t. \end{equation*}

What is the daily percent growth rate?
Rewrite $D(t)$ in the form $D(t) = 10 e^{kt}$ and find the continuous daily growth rate.

Solution.

The daily growth rate $r$ is given by $r = b-1 = 0.74 - 1 = -0.26\text{,}$ or $-26\%\text{.}$ It is negative since $D(t)$ is decreasing.
The continuous growth rate is $k = \ln (b) = \ln(0.74) \approx -0.301$ or $-30.1\%\text{.}$ Using this, then $D(t)$ in terms of the natural base is

\begin{equation*} D(t) = 10 e^{-0.301t}. \end{equation*}

Subsection Calculating the Doubling Time and the Half-Life

With logarithms, we can calculate algebraically the doubling time and the half-life of exponentially increasing or decreasing quantities.

Example 6.3.4.

You deposit $\$2,000$ into a savings account that pays $4.5\%$ annually. Let $B(t)$ denote your balance after $t$ years, which is described by

\begin{equation*} B(t) = 2000 \cdot 1.045^t. \end{equation*}

How long will it take for your money to double?
Suppose you deposit $\$5,000$ instead of $\$2,000\text{.}$ How long will it take for your money to double?

Solution.

To find the doubling time we have to solve the following equation for $t\text{:}$

\begin{equation*} 2 \cdot 2000 = 2000 \cdot 1.045^t. \end{equation*}

Divide both sides of the equation by $2000\text{:}$

\begin{equation*} 2 = 1.045^t. \end{equation*}

The unknown is in the exponent, so either the common or the natural logarithm must be applied to both sides. Using the natural logarithm (as is customary in most applied sciences) gives us the following:

\begin{align*} \ln(2) \amp= \ln(1.045^t)\\ \ln(2) \amp= t \ln(1.045)\\ t \amp= \frac{\ln(2)}{\ln(1.045)} \approx 15.75 \end{align*}

It will take approximately $15.75$ years for your money to double.
If the initial deposit is $\$5,000$ instead of $\$2,000\text{,}$ then $B(t) = 5000 \cdot 1.045^t\text{.}$ To find the doubling time of this function, solve the equation

\begin{equation*} 2 \cdot 5000 = 5000 \cdot 1.045^t. \end{equation*}

Dividing both sides by $5000$ gives the familiar equation

\begin{equation*} 2 = 1.045^t. \end{equation*}

This is the same equation as in the second step in part 1, so the solution is the same as before: $t = \ln(2) / \ln(1.045)\text{.}$ The doubling time does not depend on the initial amount—it depends only on the growth factor! That is, the initial amount is not needed to find the doubling time of an exponential function.

Example 6.3.5.

As we saw in Example 6.3.3, the amount of Diazepam, $D(t)\text{,}$ in mg, left in a patient’s system $t$ days after a $10$ mg dose is

\begin{equation*} D(t) = 10 e^{-0.301t}. \end{equation*}

Find the half-life of Diazepam.
How long will it take for the initial amount of Diazepam to be reduced to $3$ mg?

Solution.

To find the half-life, solve the following equation for $t\text{:}$

\begin{equation*} 10 e^{-0.301t} = \frac{1}{2} \cdot 10. \end{equation*}

Upon dividing by both sides by the initial value $10\text{,}$

\begin{equation*} e^{-0.301t} = \frac{1}{2}. \end{equation*}

Just as the doubling time, the half-life does not depend on the initial amount. Applying the natural logarithm to both sides of the equation yields the following:

\begin{align*} \ln(e^{-0.301t}) \amp = \ln \left(\frac{1}{2} \right)\\ -0.301 t \amp = \ln\left(\frac{1}{2}\right)\\ t \amp = \frac{\ln(1/2)}{-0.301} \approx 2.3. \end{align*}

The half-life of Diazepam is approximately $2.3$ days—it will take about $2.3$ days for the half of the initial dose to be eliminated from the body.
We have to find $t$ for which $D(t) = 3\text{,}$ so we have to solve the equation:

\begin{equation*} 10 e^{-0.301t} = 3. \end{equation*}

Just like before, divide both sides by $10\text{,}$ apply the natural logarithm to both sides, and use properties of the natural logarithm to obtain:

\begin{align*} e^{-0.301t} \amp= \frac{3}{10}\\ \ln(e^{-0.301t}) \amp = \ln(\frac{3}{10})\\ -0.301t \amp= \ln(\frac{3}{10})\\ t \amp= \frac{\ln(\frac{3}{10})}{-0.301} \approx 4. \end{align*}

It will take about $4$ days for $3$ mg to be left in a patient’s system from the initial dose of $10$ mg.

Does this answer make sense? Let’s compare it to the half-life, which is $2.3$ days. This half-life means that $10$ mg will be reduced to $5$ mg after $2.3$ days and then the remaining $5$ mg will be reduced to $2.5$ mg after the next $2.3$ days. So, it seems reasonable that $3$ mg is left after $4$ days.

Of course, you can check your answer by substituting $t = 4$ into $D(t)\text{:}$

\begin{equation*} D(4) = 10 e^{-0.301 \cdot 4} = 2.9999 \approx 3. \end{equation*}

By the way...

Example 6.3.6.

During the 1986 Chernobyl disaster, radioactive Strontium-90 was released into the atmosphere

https://semspub.epa.gov/work/HQ/175430.pdf, accessed: 5/24/2020

. Strontium-90 contaminated the nearby region and accumulated in people’s bones. (Incidentally, the isotope is often referred to as a “bone seeker”.) The half-life of Strontium-90 is $29$ years.

Find the percentage of the original amount of Strontium-90 absorbed that is still left in people’s bones in 2020.
How many years will it take for $10\%$ of the original amount of Strontium-90 to be left?

Solution.

Let $S(t)$ be the amount of Strontium-90 remaining in an affected person’s bones $t$ years after the disaster. Denote by $S_0$ the initial amount absorbed. The goal is to find a formula for $S(t)$ in terms of the natural base:

\begin{equation*} S(t) = S_0 e^{kt}. \end{equation*}

To find $k\text{,}$ we use the half-life and set up an equation for $k\text{.}$ At $t = 29\text{,}$ half of $S_0$ is left; that is, $S(29) = \frac{1}{2} \cdot S_0\text{:}$

\begin{equation*} S_0 e^{k \cdot 29} = \frac{1}{2} \cdot S_0. \end{equation*}

Divide both sides by $S_0$ so that

\begin{equation*} e^{k \cdot 29} = \frac{1}{2}. \end{equation*}

Use the natural logarithm to solve for $k\text{:}$

\begin{align*} \ln(e^{k \cdot 29}) \amp= \ln \left(\frac{1}{2}\right)\\ k \cdot 29 \amp= \ln\left(\frac{1}{2}\right)\\ k \amp= \frac{\ln(\frac{1}{2})}{29} \approx -0.0239. \end{align*}

Given this value of $k\text{,}$ a formula for $S(t)$ is

\begin{equation*} S(t) = S_0 e^{-0.0239t}. \end{equation*}

To answer the question, all that is left is to evaluate $S$ at $t = 34\text{,}$ which corresponds to the year 2020. That is,

\begin{equation*} S(34) = S_0 e^{-0.0239 \cdot 34} = S_0 \cdot 0.4437 \end{equation*}

In 2020, about $44\%$ of the original amount absorbed remains in people’s bones. Note that the percentage of Strontium-90 left does not depend on the initial amount $S_0\text{.}$
We are looking for $t$ such that $S(t) = 0.10 \cdot S_0\text{.}$ That is, for solutions to

\begin{equation*} S_0 e^{-0.0239t} = 0.10 \cdot S_0. \end{equation*}

As before, divide both sides by $S_0\text{,}$ apply the natural logarithm and use properties of the logarithm to obtain

\begin{equation*} t = \frac{\ln(0.10)}{-0.0239} \approx 96.34. \end{equation*}

Therefore, it will take approximately $96$ years before $10\%$ of the original amount remains.

Subsection The Acidity of a Liquid

In chemistry, the acidity of a liquid is measured on the $\text{pH}$ scale

http://chemistry.elmhurst.edu/vchembook/184ph.html, accessed: 5/25/2020

. The acidity depends on the hydrogen ion concentration in the liquid, denoted by $[\text{H}^{+}]$ and measured in $\dfrac{\text{mol}}{\text{L}}$ (moles per liter). The $\text{pH}$ is defined as:

\begin{equation*} \text{pH} = -\log[\text{H}^+] \end{equation*}

(The logarithm in the formula is the common logarithm.)

Example 6.3.7.

The hydrogen ion concentration of lemon juice is $[\text{H}^+] = 0.01$ $\dfrac{\text{mol}}{\text{L}}\text{.}$ Find the $\text{pH}$ of lemon juice.

Solution.

According to the $\text{pH}$ formula above, the $\text{pH}$ of lemon juice is

\begin{equation*} \text{pH} = -\log (0.01) = -\log (10^{-2}) = -(-2) = 2. \end{equation*}

Example 6.3.8.

The $\text{pH}$ of gastric acid is $1\text{.}$ Find the hydrogen ion concentration $[\text{H}^{+}]$ in gastric acid.

Solution.

To find the hydrogen ion concentration, solve the equation $1 = -\log[\text{H}^+]\text{.}$ Put another way, $-1 = \log[\text{H}^+]\text{.}$ By the definition of the common logarithm, this means

\begin{equation*} 10^{-1} = [\text{H}^+] \end{equation*}

and therefore $[\text{H}^+] = 0.1$ $\dfrac{\text{mol}}{\text{L}}\text{.}$

Exercises Exercises

Rewriting Exponential Functions.

For each of the following, rewrite a given exponential function $y=A\cdot b^t$ in terms of the natural base; that is, in the form $y=A\cdot e^{kt}\text{.}$ Round off $k$ to four decimal places.

1.

$\displaystyle y=208(0.85)^t$

Solution.

$y=208e^{-0.1625t}$

2.

$\displaystyle y=(1.09)^t$

Solution.

$y=e^{0.0862t}$

3.

$\displaystyle y= 700(1.32)^t$

Solution.

$y=700e^{0.2776t}$

4.

$\displaystyle y=2000(0.91)^t$

Solution.

$y=2000e^{-0.0943t}$

5.

If a bank offers an interest rate of $6.1\%$ with interest compounded continuously (rather than once a year), your balance $B(t)$ after $t$ years is:

\begin{equation*} B(t)=B_0e^{0.061t} \end{equation*}

where $B_0$ is your initial deposit, in dollars.

(a)

How long will it take for your money to double?

Solution.

approximately $11.36$ years

(b)

How long will it take for your money to triple?

Solution.

approximately $18.01$ years

6.

A laboratory culture of Salmonella enterica starts with $S_0$ bacteria at $t=0$ and doubles every $30$ minutes.

https://www.ncbi.nlm.nih.gov/pmc/articles/PMC6015860/, accessed: 6/26/20

Let $S(t)$ be the number of bacteria in the culture at time $t\text{,}$ in minutes.

(a)

Write a formula for $S(t)$ in terms of the natural base $e\text{.}$

Solution.

$S(t)=S_0e^{0.0231t}$

(b)

How long will it take for $S(t)$ to triple? Does the tripling time depend on the initial amount $S_0\text{?}$

Solution.

approximately $47.6$ minutes; no

7.

Bacterial population dynamics is not as simple as measuring the doubling time of a bacterium in the laboratory, under optimal growth conditions. A more challenging question is to find the doubling time of a bacterium in its natural environment, for example, in the gut. The doubling time of Salmonella enterica in the gut is $25$ hours

⁴

https://www.ncbi.nlm.nih.gov/pmc/articles/PMC6015860/, accessed: 6/26/20

. Let $G_0$ be an initial amount of Salmonella enterica in a patient’s gut at $t=0\text{.}$ Let $G(t)$ be the amount $t$ hours later.

(a)

Find a formula for $G(t)$ in base-$e$ form.

Solution.

$G(t)=G_0e^{0.0277t}$

(b)

How long will it take for the bacteria to reach $150\%$ of the initial amount?

Solution.

approximately $14.64$ hours

8.

Let $N=N(t)$ be the amount of nicotine, in milligrams, in the bloodstream of a person after a cigarette is smoked. Time $t$ is measured in hours. As with most drugs, the process of elimination of nicotine from the body is a process of exponential decay. The half-life of nicotine is $2$ hours and the amount of nicotine absorbed from a cigarette is $2$ mg.

(a)

Find a formula in the form $N(t)=A\cdot e^{kt}$ for the function $N(t)\text{.}$

Solution.

$N(t)=2e^{-0.3466t}$

(b)

How long will it take for the initial amount of nicotine to be reduced to $0.7$ mg?

Solution.

approximately $3.03$ hours

9.

Let $C(t)$ be the amount of caffeine that remains in the person’s body $t$ hours after finishing a cup of coffee. The amount of caffeine absorbed from a cup of coffee is $96$ mg. The continuous growth rate of caffeine in the body is $-13.86\%\text{.}$

(a)

Find a formula for $C(t)$ in terms of the natural base.

Solution.

$C(t)=96e^{-0.1386t}$

(b)

Find the half-life of caffeine in the body. Round off your answer to two decimal places. Include units with your answer.

Solution.

approximately $5$ hours

(c)

How long will it take for $C(t)$ to be reduced to $30\%$ of the initial amount? Does the answer depend on the initial amount?

Solution.

approximately $8.69$ hours; no

(d)

How long will it take for $C(t)$ to be reduced to $10$ mg? Does the answer depend on the initial amount?

Solution.

approximately $16.32$ hours; yes

10.

The value of an antique chair, $V(t)\text{,}$ $t$ years after it was purchased for $\$7000$ increases by $15\%$ per year.

(a)

Write a formula for $V(t)$ in the form $V(t)=A\cdot b^t\text{.}$

Solution.

$V(t)=7000(1.15)^t$

(b)

When will the value reach $\$16000\text{?}$

Solution.

approximately $5.9$ years

11.

The $\text{pH}$ of orange juice is $3\text{.}$ Find the hydrogen ion concentration $[\text{H}^{+}]$ in orange juice.

Solution.

$0.001$ $\dfrac{\text{mol}}{\text{L}}$

12.

The hydrogen ion concentration $[\text{H}^{+}]$ in tomato juice is $0.0001$ moles per liter. Find the $\text{pH}$ of tomato juice.

Solution.

The pH is $4\text{.}$

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Section 6.3 Logarithms in Applications

Objectives

Subsection Converting Between \(Q = A \cdot b^t\) and \(Q = A \cdot e^{kt}\)

Result 6.3.1. Converting An Exponential Function To And From Base \(e\).

Example 6.3.2.

Example 6.3.3.

Subsection Calculating the Doubling Time and the Half-Life

Example 6.3.4.

Example 6.3.5.

By the way...

Example 6.3.6.

Subsection The Acidity of a Liquid

Example 6.3.7.

Example 6.3.8.

Exercises Exercises

Rewriting Exponential Functions.

1.

2.

3.

4.

5.

(a)

(b)

6.

(a)

(b)

7.

(a)

(b)

8.

(a)

(b)

9.

(a)

(b)

(c)

(d)

10.

(a)

(b)

11.

12.