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How can we estimate logarithms with different bases? Take $\log_2 10$ ($1\over\log_{10}2$$\approx3.32192809$) for example. If we convert $10$ to binary, we get $1010_2$. So $\log_21010_2$ can clearly be estimated at between $3$ and $4$ because $\log_21000_2 = 3$ because $1000_2 = 8_{10}$.

How can we estimate the decimal part?

Edit: The decimal part can be estimated by seeing how close it is to the integer part. See following examples.

More examples:

$\log_3 10_{10} = \log_3101_3 \approx 2 $

$\log_4 10_{10} = \log_422_4 \approx 1.5 $

$\log_510_{10}=\log_520_5\approx1.5$

$\log_610_{10}=\log_614_6\approx1.25$

$\log_{11}10_{10}=\log_{11}A_{11}\approx1$

$\log_{11}100_{10}=\log_{11}91_{11}\approx1.9$

Calculating $\log_21010_2$:

$1010 \to 101.0 \to 10.10 \to 1.010$. Because we did it 3 times, $\log_21010_2\approx3$. Because $.01\approx1/4$ of the way to $1$, $\log_21010_2\approx3.25$

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It depends on what you have memorized. For $\log_2 10$, many people know that $\log_{10} 2\approx 0.30103$, so $\log_2 10 = \frac 1{0.30103}=\frac {10}{3}-\frac 13\% \approx 3.32$ where the last $\frac 13\%$ is optional and accounts for the difference between $0.3$ and $0.301$. The correct value is about $0.3322$

Many base $2$ logs are easier because you might know some powers of $2$

For $\log_3 10$ we know $3^2=9$ so it should be a little more than $2$. I think my next step would be to say $3^{2.5}=9*1.732=17.32-1.73=15.59$ As $10$ is about $\frac 17$ of the way from $9$ to $16$ I would add $\frac {0.5}7=0.07$ and say $\log_3 10 \approx 2.07$ The correct value is about $2.10$ Not too bad for a long linear interpolation.

I did play fair and not check the correct result until afterward. These are fairly hard as estimates go. Multiplication and division are much easier as one is more prone to have useful facts memorized. $\ln 10=2.30$ and $\ln 2=0.693$ can be useful, too.

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  • $\begingroup$ We can't do it similar to my method? With changing the number system to match the base of the logarithm? That would involve less memorizing. $\endgroup$ – Justin Apr 24 '13 at 23:32
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    $\begingroup$ @gangqinlaohu: you can do that to get the integer part just fine, if you know the applicable powers. So to get $\log_3 230$, if you know $3^5=243$ you know it is a little less than $5$. These were the things that came to my mind. If you want to do very much of this it might be useful to remember that $\frac d{dx} \log_a x=\frac 1{x \ln a}$ I have to derive that when needed. That plus knowing some natural logs and $\ln(1+x)\approx x$ will take you a long ways. For your example of $\log_2 10=3+\log_2 1.25\approx 3+0.25*\frac 1{\ln 2}\approx 3+0.25*1.4\approx 3.35$ $\endgroup$ – Ross Millikan Apr 24 '13 at 23:47
  • $\begingroup$ +1 for $\log(1+x)\simeq x$ for small $x$, that is a gem for the OP's purpose. $\endgroup$ – Lord Soth Apr 24 '13 at 23:50
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What I do is consider instead $\log_2 10^3$ (for example), if I wish to calculate $\log_2 10$. The number $2$ has the nice property that $2^{10}$ is "roughly" $1000$, which makes $$\log_2 10 = \frac{1}{3}\log_2 10^3 \simeq \frac{1}{3}\log_2 2^{10} = \frac{10}{3} \simeq 3.3.$$ I also have the unhealthy habit of calculating powers of $2$ from time to time in my head, $1024,2048,4096,8192...$

For powers of $3$, the same strategy holds $3^{21} \simeq 10^{10}$, which gives us $\log_3 10\simeq 2.1$. The whole point of these arguments is that one can get rid of the decimal parts for a while (for mental calculation purposes).

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  • $\begingroup$ what about $\log_3$? $\endgroup$ – Justin Apr 24 '13 at 23:28
  • $\begingroup$ Edited.${}{}{}{}{}$ $\endgroup$ – Lord Soth Apr 24 '13 at 23:33
  • $\begingroup$ what about $\log_n$? $\endgroup$ – Justin Apr 24 '13 at 23:34
  • $\begingroup$ Well the same strategy holds for any base, but mental calculation of the powers of that base become more and more difficult obviously. You may consult to professional mental calculation websites (or people), it is quite possible that they use different and more sophisticated strategies. $\endgroup$ – Lord Soth Apr 24 '13 at 23:46
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It's not my intent really to revive dead threads somehow, but I only recently stumble on thinking about the log in other ways than usual and I think I can provide an alternative, but useful view on it.

For more details, please see references in my blog as well as these other threads here in math.stackexchange this and this.

How to directly estimate any (well, you know...almost any) logarithm in any base?

Answer: In order to estimate $\log_b x$, estimate the number of times, $y$, that $x$ can be repeatedly divided/multiplied by $b$ until you get $1$.

It might be easier/more precise to estimate first $\log_bx^n$ and divide the result by $n$ instead.

Examples:

1) Let's estimate $\log_{57}345197$

For the sake of speeding up the estimate, let's say the base was $60$. Now $60^3\sim 216000$, so the log base 60 would be way closer to $3$ than to $4$. $57$ being smaller than $60$, we estimate it is $\log_{57}345197\sim3.0-3.5$.

The actual value up to two decimals is $3.15$, which is an error of about $1/30$. Not bad.

2) Estimate $\log_{0.57}3$. Let's consider $\log_{0.5}81\sim -(6.0-6.5)\sim - 6.25$. The sign means we are multiplying here (see the links). Hence, we can estimate to be $\log_{0.57}3$ to be somewhat larger (in absolute value) thus $\log{0.57}3\sim -6.5/4\sim -1.63$. The exact value up to two decimals is $1.95$, about $4/20=20\%$ error this time.

With a little more practice, I reckon one could make faster and more accurate estimates.

Notice that this approach really doesn't require you to memorize any log in any base: You just estimate them all over again if needed; the overhead isn't that much.

Those links provide more examples and a discussion of further topics exploring this view on its own (i.e., with only an indirect reference to exponentials).

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