In an effort to finally get to grips with logarithms (I had one related post today) I am looking for "real life" applications. Here is what I have found, and it astonishes me - they say that logarithms are great for multiplying two "big" numbers. Of course, for now let's pretend that we can't use calculator/computer to do the job. They suggest that multiplying 734 times 213 is much easier with log table, then, say, doing a straightforward multiplication on paper, or in the mind, even though the log usage doesn't give you a correct result.

link to the article

Please, help me to understand why logs are more useful. Is it only because adding together smaller numbers "sounds" easier?

Also I will be truly grateful for your help on finding meaningful, mathematically correct real life applications of logarithms. Thank you very much!

  • $\begingroup$ I used logarithm tables in high school, and we would never have used them to multiply 734 and 213, unless as an illustrative example. Now if you had 4 or 5 3-digit numbers to multiply, then you'd probably use tables (but you'd only get about a 3-significant digit approximation). In math classes they were almost exclusively used in trigonometry when solving for angles or sides of triangles, especially when the law of cosines or law of sines is used. Outside of that (and in solving exponential equations), you didn't use logarithms in math classes. $\endgroup$ May 25 '17 at 16:54
  • $\begingroup$ @Dave L. Renfro Thank you. Interesting. I will search for the usage in trigonometry. $\endgroup$
    – Vitale
    May 25 '17 at 17:23
  • 2
    $\begingroup$ Look at p. 41 of the Schaum's Outline for Trig (1954); p. 119 of Granville's Plane and Spherical ... (1909); Trotter's Manual of Logarithms ... (1841); p. 101 of Bettinger/Englund's Algebra And Trigonometry (1960); p. 85 of Howe's Mathematics for the Practical Man ... (1912) (this seems the best to begin with). $\endgroup$ May 25 '17 at 18:30
  • $\begingroup$ Especially if you go on to do calculus, you're going to find logarithms everywhere. Logarithms are one of the best functions you'll ever work with. $\endgroup$
    – Kaynex
    May 25 '17 at 18:36
  • $\begingroup$ @Dave L. Benfro Thank you very much for these suggestions! :-) $\endgroup$
    – Vitale
    May 26 '17 at 6:17

What's 128 times 512?

For those of us who grew up in the digital age and so have memorized the table of powers of 2 to many exponents (I know them to exponent 16 off top of my head), this is easy using logarithms base 2: $$128 * 512 = 2^7 * 2^9 = 2^{16} = 65536 $$ Notice: this is nothing more nor less than a logarithm calculation. Let me lay it out in a different way. $$\log_2(128) = 7 $$ $$\log_2(512) = 9 $$ $$\log_2(128 * 512) = \log_2(128) + \log_2(512) = 7 + 9 = 16 $$ So, $$128 * 512 = 2^{16} = 65536 $$ At all stages, I have consulted a memory device (in this case, my brain) to do the calculations.

Okay, now let me turn to your problem. $$734 * 213 $$ You say "the log usage doesn't give the correct result". Well, if you want all 6 digits of the answer to be completely correct, that's true.

But wait, how did I even know that the answer was 6 digits? I knew this because, in my head and very rapidly, I did an estimated calculation that amounts to the following logarithm calculation (it takes me 50 times longer to write this out than it took to do it in my head): \begin{align*} \log_{10}(734) &= \log_{10}(100 * 7.34) = \log_{10}(100) + \log_{10}(7.34) = 2 + \log_{10}(7.34) \\ \log_{10}(213) &= \log_{10}(100 * 2.13) = \log_{10}(100) + \log_{10}(2.13) = 2 + \log_{10}(2.13)\\ \log_{10}(734 * 213) &= \log_{10}(734) + \log_{10}(213) = 2 + \log_{10}(7.34) + 2 + \log_{10}(2.13) \\ &= 4 + \log_{10}(7.34) + \log_{10}(2.13) \\ &= 4 + \log_{10}(7.34 * 2.13) \end{align*} Now I apply this to do an estimate. $$4 + \log_{10}(7 * 2) < 4 + \log_{10}(7.34 * 2.13) < 4 + \log_{10}(8*3) $$ $$4 + \log_{10}(14) < \log_{10}(734 * 213) < 4 + \log_{10}(24) $$ $$\log_{10}(10000*14) < \log_{10}(734 * 213) < \log_{10}(10000*24) $$ $$\log_{10}(140000) < \log_{10}(734 * 213) < \log_{10}(240000) $$ and so I conclude that $$140000 < 734 * 213 < 240000 $$ Hence I know that $734 * 213$ is a six digit number, in fact I know quite a bit more.

This is the real value of logarithms: it lets you do estimated calculations very, very quickly. The better logarithm table you have, the better and more accurate your calculation will be.

  • $\begingroup$ Thanks for the edits, and so much for my arithmetic abilities! $\endgroup$
    – Lee Mosher
    May 25 '17 at 17:11
  • $\begingroup$ Thank you for your help, and explanation. I am reading the rest of your post, and try to recall log rules. Much appreciated. Learned a lot. $\endgroup$
    – Vitale
    May 25 '17 at 17:12
  • $\begingroup$ To avoid jumping in with edits this time, I decided to ask: isn't x in $$\log_{10} (x) = 4$$ equals 10000, and not 100000? $\endgroup$
    – Vitale
    May 25 '17 at 17:19
  • $\begingroup$ Yes that's right, but the extra digit comes from the fact that $7*2=14$ has two digits. $\endgroup$
    – Lee Mosher
    May 25 '17 at 17:36
  • $\begingroup$ I will add another line or two to the calculation to emphasize this point better. $\endgroup$
    – Lee Mosher
    May 25 '17 at 17:40

The practical real-life application of logarithms in doing arithmetic came before the days of hand-held electronic calculators (if you are young, let me explain that a hand-held calculator was your cell phone plus its calculator app, minus the ability to text, take pictures, access the internet, spend money on other apps, or -gulp- talk on the phone).

Engineers would carry a slide rule, which is a pair of sticks with logarithmic-spaced rulings marked on them. You could use a slide rule to multiply, say, $743\times 213$ in less than a second, getting the answer to three-digit accuracy.

As to on-paper calculations, although the theoretical complexity of multiplication of $b$-bit numbers is the same as that of addition which is $O(b\log b)$, divide-and-conquer algorithms are needed to achieve this efficiency, and they are too complicated to be practical by hand.

The usual on-paper multiplication takes $O(b^2)$ work, whle looking up logs and adding them is again $O(b\log b)$ so there can be a practical advantage.


Yes, you got it correctly: adding two small numbers is much easier and faster than multiplying two big ones. Obviously nowadays it's not very likely that you have at hand a log table instead of a calculator (e.g. within the smartphone) so the most important real life applications of logs are others.


enter image description here

Above is a table of logarithms - such as would have occurred in the back of most high school Algebra II books.

This is how I would have used it to compute $345 \times 0.0582$

Look down the "N" column for "34" and intersect that row with the column labelled $5$. You now know that $\log 3.45 = 0.5378$. Hence $$\log 345 = 2.5378$$

Similarly $\log 5.82= 0.7649$. In scientific notation, $0.0582 = 5.82 \times 10^{-2}$. For better or for worse, negative logarithms were avoided. Hence we wrote $$\log 0.0582 = 8.7649 - 10$$

No, really!. Adding the logarithms, we get

\begin{array}{rrrr} 2.5378 \\ + 8.7649 & - 10 \\ \hline 11.3027 &- 10 \end{array}

Which simplifies to $$\log(345 \times 0.0582) = 1.3027$$

As most always happens, $3027$ is not in the table. We will need to do some linear interpolation. We make the following table

\begin{array}{|c|c|} \hline 2.000 & .3010 \\ x & .3027 \\ 2.010 & .3032 \\ \hline \end{array}

Ignoring the decimal points and superfluous digits, we compute

\begin{align} \dfrac{x-00}{10-00} &= \dfrac{27-10}{32-10} \\ \dfrac{x}{10} &= \dfrac{17}{22} \\ x &= \dfrac{170}{22} \\ x &= 8 \end{align}

Hence $$\log 2.008 = .3027$$

We conclude $$345 \times 0.0582 = 20.08$$

In fact, $345 \times 0.0582 = 20.079$.


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