# Decimal Approximations of Logarithms (Part 2)

While some common (i.e., base-10) logarithms work out evenly, like $\log_{10} 10,000$, most do not. Here is the typical output when a scientific calculator computes a logarithm: To a student first learning logarithms, the answer is just an apparently random jumble of digits; indeed, it can proven that the answer is irrational. With a little prompting, a teacher can get his/her students wondering about how people 50 years ago could have figured this out without a calculator. This leads to a natural pedagogical question:

Can good Algebra II students, using only the tools at their disposal, understand how decimal expansions of base-10 logarithms could have been found before computers were invented?

Here’s a trial-and-error technique — an exploration activity — that is within the grasp of Algebra II students. It’s simple to understand; it’s just a lot of work. While I don’t have a specific reference, I’d be stunned if none of our ancestors tried something along these lines in the years between the discovery of logarithms (1614) and calculus (1666 or 1684).

To approximate $\log_{10} x$, look for integer powers of $x$ that are close to powers of 10.

I’ll illustrate this idea with $\log_{10} 3$. $3^1 = 3$ $3^2 = 9$

Not bad… already, we’ve come across a power of 3 that’s decently close to a power of 10. We see that $3^2 = 9 < 10^1$

and therefore $\log_{10} 3^2 < 1$ $2 \log_{10} 3< 1$ $\log_{10} 3< \displaystyle \frac{1}{2} = 0.5$ Let’s keep going. We just keep multiplying by 3 until we find something close to a power of 10. In principle, these calculations could be done by hand, but Algebra II students can speed things up a bit by using their scientific calculators. $3^3 = 27$ $3^4 = 81$ $3^5 = 243$ $3^6 = 729$ $3^7 = 2,187$ $3^8 = 6,561$ $3^9 = 19,683$ $3^{10} = 59,049$ $3^{11} = 177,147$ $3^{12} = 531,441$ $3^{13} = 1,594,323$ $3^{14} = 4,782,969$ $3^{15} = 14,348,907$ $3^{16} = 43,046,721$ $3^{17} = 129,140,163$ $3^{18} = 387,420,489$ $3^{19} = 1,162,261,467$ $3^{20} = 3,486,784,401$ $3^{21} = 10,460,353,203$

This looks pretty good too. (Students using a standard ten-digit scientific calculator, of course, won’t be able to see all 11 digits.) We see that $3^{21} > 10^{10}$

and therefore $\log_{10} 3^{21} > \log_{10} 10^{10}$ $21 \log_{10} 3 > 10$ $\log_{10} 3 > \displaystyle \frac{10}{21} = 0.476190\dots$

Summarizing our work so far, we have $0.476190\dots < \log_{10} 3 < 0.5$.

We also note that this latest approximation actually gives the first two digits in the decimal expansion of $\log_{10} 3$. To get a better approximation of $\log_{10} 3$, we keep going. I wouldn’t blame Algebra II students a bit if they use their scientific calculators for these computations — but, ideally, they should realize that these calculations could be done by hand by someone very persistent. $3^{22} = 31,381,059,609$ $3^{23} = 94,143,178,827$ $3^{24} = 282,429,536,481$ $3^{25} = 847,288,609,443$ $3^{26} = 2,541,865,828,329$ $3^{27} = 7,625,597,484,987$ $3^{28} = 22,876,792,454,961$ $3^{29} = 68,630,377,364,883$ $3^{30} = 205,891,132,094,649$ $3^{31} = 617,673,396,283,947$ $3^{32} = 1,853,020,188,851,841$ $3^{33} = 5,559,060,566,555,523$ $3^{34} = 16,677,181,699,666,569$ $3^{35} = 50,031,545,098,999,707$ $3^{36} = 150,094,635,296,999,121$ $3^{37} = 450,283,905,890,997,363$ $3^{38} = 1,350,851,717,672,992,089$ $3^{39} = 4,052,555,153,018,976,267$ $3^{40} = 12,157,665,459,056,928,801$ $3^{41} = 36,472,996,377,170,786,403$ $3^{42} = 109,418,989,131,512,359,209$ $3^{43} = 328,256,967,394,537,077,627$ $3^{44} = 984,770,902,183,611,232,881$

Using this last line, we obtain $3^{44} < 10^{21}$

and therefore $\log_{10} 3^{44} < \log_{10} 10^{21}$ $44 \log_{10} 3 < 21$ $\log_{10} 3 < \displaystyle \frac{21}{44} = 0.477273\dots$

Summarizing our work so far, we have $0.476190\dots < \log_{10} 3 < 0.477273\dots$.

A quick check with a calculator shows that $\log_{10} 3 = 0.477121\dots$. In other words,

• This technique actually works!
• This last approximation of $0.477273\dots$ actually produced the first three decimal places of the correct answer! With a little more work, the approximations $3^{109} \approx 1.014417574 \times 10^{52} > 10^{52}$ $3^{153} \approx 9.989689095 \times 10^{72} < 10^{73}$

can be found, yielding the tighter inequalities $\displaystyle \frac{52}{109} < \log_{10} 3 < \displaystyle \frac{73}{153}$,

or $0.477064\dots < \log_{10} 3 < 0.477124$.

Now we’re really getting close… the last approximation is accurate to five decimal places.

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