Explaining Logarithm

What is logarithm?

The logarithm of a number is -more or less- how many digits it has.

So the logarithm of \(8347238\) is about \(7\) (actually it's \(6.921....\)). The logarithm of \(9\) is about \(1\). So it is a convenient way to measure the scale of a number.

Now this was a very rough idea. Let's make it more precise. If you take any number and multiply it by \(10\),it gains an additional digit, so according to what I said above, the logarithm increases by one. So what actually happens is, the logarithm of a number is how often I have to multiply \(10\) to get that number. So \(\log_{(100)}=2\) because \(10 \times 10 = 100\), so I multiplied two \(10s\) together. Similarly, \(\log_{(10000)}=4\) because \(10 \times 10 \times 10 \times 10=10000\), and so on. Multiplying four \(10s\) together can also be written as \(10^4\). And now we can generalise the idea:

\(\log_{(x)}= y\) is another way of saying that \(10^y = x\).

The final question you may have is: what's so special about the number \(10\)? Why does it count how many \(10s\) I have to multiply together? Well actually, logarithm is used with other base numbers as well: base \(2\) and base \(e\) are also very common. But for each choice, you get logarithm function that measures the scale of the number you put in.

What are logarithms used for? Are decibels a good example of the usefulness of a logarithm? Are logarithms calculus?

Logarithms can be used to talk about things that can be both tiny and gigantic, such as in

- earthquake magnitudes,

- noise levels in decibels, and

- acidity (pH).

A big earthquake can be millions of times bigger than a tiny one. If you tried to make a bar graph where the bars has sizes \(10, 100,\) and \(10 000 000\), it would look stupid. The bars of size \(10\) and \(100\) would be too small to see, and you won't be able to tell that one of them is ten times bigger than the other. If you instead take the logarithm of each number, you get \(1, 2,\) and \(7.\) That makes a bar graph you can understand.

Keep that in mind when you hear about earthquake magnitudes. A \(7.0\) earthquake is ten times bigger than a \(6.0\) earthquake, which is ten times bigger than a \(5.0\) earthquake. Taking logarithms lets us put an earthquake caused by a stick of dynamite \((1.2)\) on the same scale as the \(2011\) earthquake in Japan \((9.0)\).

Logarithms can also be used to measure how long it will take something to grow exponentially or decay exponentially,such as

- money growing with a fixed interest rate,

- bacteria growing in a petri dish,

- radioactive decay, for example in Radiocarbon dating, and

- the sound made by a bell.

If you have bacteria that divide every \(30\) minutes and are currently taking up \(0.1\)% of the petri dish, you can use logarithms to estimate how long it will take them to fill up the entire dish. The same goes for \($5000\) in an account with a \(2\)% interest rate. If you leave the interest in the account, logarithms will tell you when you'll have \($6000\).

Logarithms can also be used in calculations by turning multiplication into addition. If I gave you the option between multiplying twenty numbers together by hand or adding twenty numbers together by hand, you'd pick the second option. If you need to multiply twenty numbers, you can instead take the base \(10\) logarithm of each number, add the results, and then raise \(10\) to that power. Finding the logarithm might seem hard, but, in the past, people could just look it up in a logarithm table or use a slide rule. Finding the answer using logarithms was way faster.

The problem of multiplying lots of numbers was the original reason logarithms were developed. This method is now obsolete thanks to computers, which are pretty fast at multiplying. In the meantime, though, we've discovered tons of uses of logarithms, most of which I haven't even listed here.

What is the intuition behind the logarithm?

The logarithm counts the number of groupings.

Suppose a bakery puts \(12\) cookies in a package, and places \(12\) of these packages in a larger box for transport:

Then a box can be seen as cookies which have been grouped twice: one box contains \(12^2 = 144\) cookies. Inversely, when ordering \(144\) cookies, and knowing that this bakery works with base \(12,\) the logarithm will return the number of groupings: \(\log_{12}{144} = 2\) groupings. Suppose the transport company also works in base twelve; \(12\) boxes are wrapped in plastic, \(12\) plastic units are stacked onto a wooden pallet, \(12\) pallets are transported in a van, which makes \(12^3=1728 \) boxes per ride:

Now the total number of cookies in one transport, is obtained by multiplying the number of cookies per box with the number of boxes per ride: \(144 \times 1728 = 248832\) cookies.
However, if we look at this on a grouping scale (logarithmic scale) then we must use addition instead: the total number of groupings is \(2 + 3 = 5\). Stated otherwise: \(\log_{12}{144} + \log_{12}{1728} = \log_{12}{248832}\). Now suppose \(20736\) packages of cookies were stolen. Then we know that this involved \(\log_{12}{20736} = 4\) groupings. So counting upwards from these packages...

...we know that this monster learned how to drive:

Note by Cera Mess
3 months, 3 weeks ago

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How do I setup a logarithm for calculating the increase from interest rate or that petri dish thing you mentioned? I've always wondered that.

The Strategy Gamer - 3 months, 3 weeks ago

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For the petri dish example, at

\(t=0\) the petri dish is \(0.1\)% full,

\(t=30min\) the petri dish is \(0.1\)% \(\times 2\) full,

\(t=60min\) the petri dish is \(0.1\)% \(\times 2^2\) full,

\(t=90min\) the petri dish is \(0.1\)% \(\times 2^3\) full,

\(t=120min\) the petri dish is \(0.1\)% \(\times 2^4\) full.

You can see that a general formula for how full the petri dish at time \(t\) is \(0.1\)% \(\times 2^{\left(\frac{ t }{ (30min)}\right )}\). For example, for \(t=120min, \frac{t }{(30min) }= 4,\) and indeed the petri dish is \(0.1\)% \(\times 2^4\) full.

Assuming this growth rate continues until the petri dish is full, to find when that happens, we want to find the time t so that \(0.1\)% \(\times 2^{\left(\frac{t} {(30min)}\right)} = 100\)%. We begin by dividing both sides by \(0.1\)% to get \(2^{\left( \frac{t }{ (30min) }\right)} = 1000.\) Then we take the base two logarithm of both sides. \(\frac{t}{ (30min)} = log_2(1000)\) Then we multiply both sides by \(30min\) to get \(t = (30min) \times log_2(1000)\), or about \(5\) hours.

Cera Mess - 3 months, 3 weeks ago

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Thanks! It's easier than any logarithmic equation exercise they teach us, and yet no one says anything about how useful this is in real life. Too bad high school algebra is all like that.

The Strategy Gamer - 3 months, 3 weeks ago

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Using logarithms is still a useful way to multiple numbers. While for small numbers, it's probably faster to just multiply - as the computer probably has a dedicated circuit for it - for larger numbers (numbers with a lot more bits than the word size), using logarithms to change it into an addition problem is probably going to be faster. This is especially true the less we care about the precise accuracy, and our numbers are too large to represent as floating point numbers (which has good hardware support).

Joms Joseph Leelin - 3 months, 3 weeks ago

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" This is especially true the less we care about the precise accuracy, and our numbers are too large to represent as floating point numbers (which has good hardware support)"

I didn't get it?

Jeff Ishee - 3 months, 3 weeks ago

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Floating point doubles top out around 10^300, and even quads top out around 10^5000. Doing calculations on larger numbers requires other methods.

Rafael Melo - 3 months, 3 weeks ago

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