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.
