I was studying logarithms, and had to solve the problem:

If $\log 8 = 0.90$, find $\log 0.125$.

I found out the answer to be $-0.90$. That was easy. But my text book has given the answer as:

$$-0.90 = \bar{1}.10$$

Now what that bar above $1$ is for ? My understanding is that it is used to signify a repeating digit. But how can $-0.90 = \bar{1}.10$ ?

Can someone please explain this to me.

Thank you!

  • $\begingroup$ Which text book is this? Looks like very strange notation to me... $\endgroup$
    – 5xum
    Mar 20 '14 at 9:59
  • $\begingroup$ @5xum This $\endgroup$ Mar 20 '14 at 10:01

This was a very common notation when decimal logarithms were involved. The starting point is that if $$ M/N=10^k $$ for two positive numbers $M$ and $N$, then their decimal logarithms have the same mantissa (fractional part) and we can write $$ \log_{10} M= k + \log_{10} N $$ In particular, any number $M>0$ can be uniquely written as $$ M=10^k N $$ with integer $k$ and $1\le N<10$. So the logarithmic tables were compiled, say, for numbers from $1$ to $1000$, but only showed the mantissa; when the logarithm of, say, $10.4$ was needed, one looked at $104$ finding the mantissa to be $01703$ and so could conclude $\log_{10}10.4=1.01703$. If one knows the decimal logarithm of $2$ $$ \log_{10}2=0.30103 $$ (equality is meant as “approximate”), then $\log_{10}0.02=-2+0.30103$, which was written as $$ \log_{10}0.02=\overline{2}.30103 $$ to ease lookup in logarithmic tables, because, as already said, only mantissas were shown.

In your case, $0.125=1/8$, so $$ \log_{10}0.125=-\log_{10}8 = -3\log_{10}2=-0.90309=-1+0.09691=\overline{1}.09691 $$ (use the “complement to $9$” rule for writing the mantissa).

  • $\begingroup$ where you say the equality is "approximate", you can actually use the $\LaTeX$ symbol. May I edit?(+1) $\endgroup$
    – Guy
    Mar 20 '14 at 10:12
  • $\begingroup$ @Sabyasachi I preferred the notation that was standard when doing those computations. $\endgroup$
    – egreg
    Mar 20 '14 at 10:13
  • $\begingroup$ Okay. Cheers.${}$ $\endgroup$
    – Guy
    Mar 20 '14 at 10:13
  • $\begingroup$ @egreg Wonderful answer! Thank you! But I was wondering if at the last, it should be $-0.90309$ instead of $0.90309$ ?? $\endgroup$ Mar 20 '14 at 10:16
  • 2
    $\begingroup$ @GaurangTandon just to be clear for others who may be reading this $0.9691=\log_{10} 1.25$ so the expression given by egreg is the logarithm of $1.25 \times 10^{-1}=0.125$. The notation was used because it allowed easier computation in the days before computers and calculators (and also made sense in the era of slide rules). $\endgroup$ Mar 20 '14 at 10:29

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