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Just Enough Algebra

Section 0.9 Prelude: Logarithms

Nearlywed, hellscape, bodycon, and woke. Just a few new words added to dictionaries in 2023. People love to create new words and phrases. These new words spread through social media, music, and word of mouth.
I created a new word “puzzlaxing” meaning relaxing by doing puzzles. In one week, maybe 10 people will have heard of it. After another week, perhaps 100 people. Then 1,000 the next week, and so on. How many weeks until 1 million people are have heard of my new word “puzzlaxing"?
Notice that in 1 week is \(10 = 10^1\) people, two weeks is \(10^2 = 100\) people, three weeks is \(10^3=1000\) people, and so on. We are looking for a number \(W\) where \(10^W=1{,}000{,}000\) people. Aha! \(10^6=1{,}000{,}000\) and so \(W=6\) weeks from now.
Suppose we wanted to determine when 5,000 people had heard of “puzzlaxing”. That means we want a number \(W\) where \(10^W = 5{,}000\text{.}\) Now what? The answer is somewhere between 3 and 4 weeks because \(10^3=1{,}000\) and \(10^4=10{,}000\text{.}\) That’s probably a good enough answer – between 3 and 4 weeks, but suppose we want the exact moment.
Let’s try guessing. How about 3.5 weeks?
\begin{equation*} 10^{3.5} =10 \wedge 3.5 = 3{,}162.27\ldots \end{equation*}
which is much smaller than 5,000. How about 3.7 weeks?
\begin{equation*} 10^{3.7} =10 \wedge 3.7 = 5{,}011.87.. \end{equation*}
which is slightly bigger than 5,000. I want to find the answer here so let’s try
\begin{equation*} 10^{3.69}=10 \wedge 3.69= 4{,}897.78\ldots \end{equation*}
or finally
\begin{equation*} 10^{3.699}=10 \wedge 3.699= 5{,}000.34\ldots \end{equation*}
That’s as close as I’m gonna guess.
Okay, I’m curious. Is there an exact power of 10 that gives 5,000? Your calculator should have a key that says log or maybe LOG. Try typing
\begin{equation*} \log(5000)= 3.6989700043\ldots \end{equation*}
A small note here about parentheses. Some calculators give the first parenthesis for free when you type log but you have to type the closing parenthesis in yourself.
Check it out, that’s the answer we were looking for:
\begin{equation*} 10^{3.6989700043} =10 \wedge 3.6989700043 = 4{,}999.999~999~5853\ldots \approx 5{,}000 \end{equation*}
If we had kept more digits it would have been actually 5,000.
What is this log key doing? First, “log” is short for logarithm base 10. There are other bases, but 10 is what we’ll focus on in this course. Try these calculations:
\begin{align*} \log(10) \amp =1\\ \log(100) \amp =2\\ \log(1{,}000) \amp =3\\ \log(10{,}000) \amp= 4\\ \log(100{,}000) \amp= 5\\ \log(1{,}000{,}000) \amp= 6 \end{align*}
What do you see? In each case the logarithm is the number of zeros or, equivalently, it’s the power of 10. For example, 10,000 has 4 zeros and \(10^4= 10{,}000\) and \(\log(10{,}000) = \log(10^4)=4\text{.}\) In other words, a logarithm is just an exponent. And logarithms help us find the exponent. Makes sense.
What about logs of numbers that aren’t just powers of 10? Here are some examples.
\begin{align*} \log(25) \amp = 1.3979\ldots \\ \log(250) \amp = 2.3979\ldots \\ \log(2{,}500) \amp = 3.3979\ldots \\ \log(25{,}000) \amp = 4.3979\ldots \end{align*}
To see what’s happening we want to involve powers of 10. Scientific notation will do that for us. Let’s write these numbers in scientific notation and see what we learn. For example \(25{,}000 = 2.5 \times 10^4\) and
\begin{equation*} \log( 25{,}000)=\log(2.5 \times 10 ^4)=4.3979\ldots \approx 4 \end{equation*}
Before we write down a general rule, let’s check more numbers.
\begin{equation*} \log(7{,}420{,}000) = \log(7.42 \times 10^6)=6.870403905\ldots \approx 6 \end{equation*}
\begin{equation*} \log (4)=\log(4 \times 10^0)=0.602059991\ldots \approx 0 \end{equation*}
\begin{equation*} \log (0.00917)= \log(9.17 \times 10^{-3}) = -2.037630664\ldots \approx -3 \end{equation*}
In every case we are rounding down, but it’s always the same:
log(number) \(\approx\) power of 10 in the scientific notation for that number.

Do you know …

  1. What a logarithm (base 10) means?
  2. How to evaluate logarithms (base 10) on a calculator?
  3. Which size numbers have a positive log and which have a negative log (base 10)?
  4. The connection between logarithms (base 10) and scientific notation.
If you’re not sure, work the rest of exercises and then return to these questions. Or, ask your instructor or a classmate for help.

Exercises Exercises

Exercises 1-4 are available in a separate workbook format.

1.

The number of bacteria in a dish increases 10-fold each day. Note: 10-fold means \(\times 10\text{.}\) Suppose we had 1 microliter of bacteria at the start of the first day. That means after \(T\) days there will be \(10^T\) microliters of bacteria.
(a)
How many bacteria (in microliters) will there be after 1 day? After 2 days? After 3 days?
(b)
In how many days will the bacteria have reached 1 liter, which is 1 million microliters?
(c)
How can we use logs to find the answer?

2.

The problem continues …
(a)
Approximately how many days (from the start) does it take to reach the 25 millilter capacity of the petri dish, which is 25,000 microliters? Guess and check to find the answer to 1 decimal place.
(b)
How can we use logs to find the answer?
(c)
Convert your answer to days & hours format (“ days and hours”).

3.

The equation \(pH = -\log(H^+)\) tells us the pH of a substance (on a scale from 0 to 14) based on its molar hydrogen ion concentration \(H^+\text{.}\) Don’t let the notation here scare you: \(pH\) is a single quantity and \(H^+\) has nothing to do with exponents or adding.
For example, lemon juice has \(H^+= 0.0025\) and so the pH of lemon juice is
\begin{equation*} -\log(0.0025) = \text{(-)} \log(0.0025) = 2.6020599913 \approx 2.6 \end{equation*}
(a)
Coca-Cola has \(H^+= 0.000~398\text{.}\) Find the pH of Coca-Cola. Note: the funny spaces are to help you read the number.
(b)
Hair shampoo has \(H^+= 0.000~003~162\text{.}\) Find the pH of hair shampoo.
(c)
Household bleach has \(H^+=1.1 \times 10^{-13}\text{.}\) Find the pH of bleach.
(d)
Materials with pH values between 0-5 are acidic, between 9-14 are basic, and between 5-7 are neutral. Which of the above materials are acidic, basic, and neutral?

4.

In Minneapolis, apartment rent is expected to increase by 16% next year.
(Story also appears in 0.3.2 and 0.7.3)
(a)
Astra lives in a 1-bedroom apartment where they pay $825 per month in rent. If their rent increased by 16%, in how many years would their rent be doubled to $1,650? As we’ll see later, the answer is \(\displaystyle \frac{\log(2)}{\log(1.16)}\text{.}\) Don’t forget to close the parentheses.
(b)
Lucky for Astra, their building is subject to rent stabilization laws and so their rent cannot increase by more than 3%. In how many years would their rent double under this cap? The answer is \(\displaystyle \frac{\log(2)}{\log(1.03)}\text{.}\)

5.

According to our story, in approximately how many weeks will 30,000 people have heard of “puzzlaxing”?
(a)
Since 30,000 is between 10,000 and 100,000, what does that tell us about the answer?
(b)
Guess to try to find the answer, the number \(W\) where \(10^W = 30{,}000\text{.}\) It’s okay to get the answer to one decimal place.
(c)
Use logs to find an exact answer.

6.

The intensity of a sound in decibels is calculated using a logarithm. For example, a sound 100 times the level humans can hear has an intensity of
\begin{equation*} 10\log (100)= 10 \times \log(100)= 20\text{ decibels} \end{equation*}
(a)
Calculate the intensity, in decibels, of a cat purring which averages about 300 times the level humans can hear using the formula \(10\log(300)\text{.}\)
(b)
Calculate the intensity, in decibels, of normal conversation which averages about 1 million times the level humans can hear using the formula \(10\log(1{,}000{,}000)\text{.}\)
(c)
Many young adults are at high risk of hearing loss because they crank the volume of music they’re listening to, often to 2 trillion times the level humans can hear. Calculate that intensity in decimals using the formula \(10\log(2 \ast10^{12})\text{.}\) Prolonged sound above 70 decibels may damage your hearing and any loud noise above 120 decibels can cause immediate harm to hearing.

7.

In 2021, Arrietty charged $5,000 on her credit card to help pay tuition. Her card charges interest at the rate of 20.7% APR. We are going to ignore any minimum payments or fees.
(a)
Arrietty was hoping to pay the debt back quickly, but in 2023 she had not paid any of the debt. Calculate the amount due on that original charge using the formula \(5000 \ast 1.207^2\text{.}\)
(b)
If she continues to leave the debt unpaid, when will her debt pass $10,000? As we’ll see later, the answer is \(\displaystyle \frac{\log(2)}{\log(1.207)}\) years after 2020. Don’t forget to the close the parentheses.

8.

Darcy likes to use temporary hair color in wild colors. Good thing it washes out. Her best guess is that 8% of the color washes out each time she shampoos her hair. That means \(100\% - 8\% = 92\%\) of the color remains after each shampoo.
(Story also appears in 3.4.9)
(a)
What percentage of the color will be remain after Darcy washes her hair three times? Calculate the percentage using the formula \(100 \ast 0.92^3\text{.}\)
(b)
After how many shampoos will half of the color be gone? As we’ll see later, the answer is \(\displaystyle \frac{\log(0.5)}{\log(0.92)}\text{.}\) Don’t forget to the close the parentheses.