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.
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!
Best Answer
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.