Approximating log107\log_{10}7

The Moderators have been talking about methods to approximate logarithms given only a little bit of information; namely, log1020.301\log_{10}2\approx0.301 and log1030.477.\log_{10}3\approx0.477. I'm going to go through a couple methods to provide estimates.

It should be noted that the true value of log107\log_{10}7 is 0.845098...0.845098...


Let's start off on a large scale. Obviously, log106<log107<log108.\log_{10}6<\log_{10}7<\log_{10}8. To find log106\log_{10}6 and log108,\log_{10}8, we use the given information. log106=log102+log1030.301+0.477=0.778.\log_{10}6=\log_{10}2+\log_{10}3\approx0.301+0.477=0.778. Also, log108=3log1023×0.301=0.903.\log_{10}8=3\log_{10}2\approx3\times0.301=0.903. So 0.778<log107<0.903.0.778<\log_{10}7<0.903. Not a very good estimate. Even if we take the average of these, we get 0.8405,0.8405, which is only accurate to two decimal places.

Let's be a little more creative in our thinking. The set of numbers {48,49,50}\{48,49,50\} includes numbers whose logarithms can be expressed with our given information and a power of 7.7. Once again, log1048<log1049log1050\log_{10}48<\log_{10}49\log_{10}50\Rightarrow log1048<2log107<log1050.\log_{10}48<2\log_{10}7<\log_{10}50. The only difference here is that 5050 has a power of 55 in its factorization, but log105\log_{10}5 can easily be eliminated by noticing that log1050=log101002=log10100log102=2log102.\log_{10}50=\log_{10}\frac{100}{2}=\log_{10}100-\log_{10}2=2-\log_{10}2.

log1048=4log102+log1031.681.\log_{10}48=4\log_{10}2+\log_{10}3\approx1.681. Also, log1050=2log1021.699.\log_{10}50=2-\log_{10}2\approx1.699. Dividing by 2,2, we find that 0.8405<log107<0.8495.0.8405<\log_{10}7<0.8495. This is better. Not to mention, taking the average of these values yields log1070.8450,\log_{10}7\approx0.8450, accurate to 44 decimal places.


What method would you use to calculate log107?\log_{10}7? Please share what you think!

Note by Trevor B.
4 years, 5 months ago

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24002401log2400log2401log(23×3×102)log74log23+log3+log1024log73log2+log3+24log7log70.8450\begin{aligned} 2400 & \approx & 2401 \\ \log 2400 & \approx & \log 2401 \\ \log (2^3 \times 3 \times 10^2) & \approx & \log 7^4 \\ \log 2^3 + \log 3 + \log 10^2 & \approx & 4 \log 7 \\ 3 \log 2 + \log 3 + 2 & \approx & 4 \log 7 \\ \log7 & \approx & 0.8450 \\ \end{aligned}

Pi Han Goh - 4 years, 5 months ago

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So, that begs the question of
1. How do I make such an amazing observation that 24002401 2400 \approx 2401 ?
2. How do I know what the next approximation should be?

Think about this some, before reading my other comment.

Calvin Lin Staff - 4 years, 5 months ago

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  1. I just found numbers that are close to powers of 77 that are can be factored in terms of 2,3,5,72,3,5,7

  2. I don't have a solid answer, but I'm thinking we should find larger pairs of numbers such that their percentage difference is as small as the previous, like: 657777=79(1041)79×1046^5 \approx 7777 = \frac {7}{9}(10^4 - 1) \approx \frac {7}{9} \times 10^4 or 67280000=104×22×76^7 \approx 280000 = 10^4 \times 2^2 \times 7 , but both methods yields 0.8440.844 as the approximation, so I'm kinda stumped right now.

In the meantime, lemme peruse your other comment. =)

Pi Han Goh - 4 years, 5 months ago

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There's a 0.04% error in approximating 2401 as 2400, but works out quite well :) +1

A Brilliant Member - 4 years, 5 months ago

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@Calvin Lin and @Daniel Liu were the moderators discussing this in our messageboard.

Trevor B. - 4 years, 5 months ago

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Cool!! So you guys have a message board of your own?

A Brilliant Member - 4 years, 5 months ago

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The general idea is that if you can bound a<7n<b a < 7 ^ n < b , then you should also look at how ab ab compares to 72n 7^{2n} . This is akin to the Root approximation - bisection method.

For example, from above, we have 6<7<8 6 < 7 < 8 , which tells us we should compare 48 48 with 49 49 , and work with 48<49<82 48 < 49 < 8^2 .

Explicitly, we can use the following series of inequalities:
Step 1: 6<7<8 6 < 7 < 8 . Comparing 6×8 6 \times 8 with 72 7^2 , we have 48<72 48 < 7^2 . Hence, this leads to:
Step 2: 48<72<82=64 48 < 7^2 < 8^2 = 64. Comparing 48×64 48 \times 64 with 74 7^4 , we have 74<3072 7^4 < 3072 . Hence, this leads to:
Step 3: 2304=482<74<3072 2304 = 48^2 < 7^4 < 3072 . Comparing 2304×3072 2304 \times 3072 with 78 7 ^ 8 , we have 78<7077888 7^ 8 < 7077888 . Hence this leads to:
Step 4: 5308416=23042<78<7077888 5308416 = 2304^2 < 7^8 < 7077888 .

We can continue this process indefinitely to bound log7 \log 7 as tightly as we want to. At step 1, we have 0.778<log7<0.903 0.778 < \log 7 < 0.903 , and at step 4 we have 0.840<log7<0.856 0. 840 < \log 7 < 0.856 .

Question: Why did I say that this is "akin to the Root approximation - bisection method"?
(see comments below for the explanation)


Note: As it turns out, we could make the observation that 50 a much better bound. With 48<49<50 48 < 49 < 50 , and so we should compare 48×50 48 \times 50 with 492 49^2 .

Calvin Lin Staff - 4 years, 5 months ago

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Because you want to find values a,ba, b such that the inequality is satisfied a<7n<ba < 7^n < b

Taking log to both sides loga<nlog7<logb \log a < n \log 7 < \log b

Bisection method implies we must minimize the difference between loga+logb2 \frac { \log a + \log b }{2} and nlog7 n \log 7 .

Set both of them to be equal implies we want to compare ab ab to 72n 7^{2n}

Pi Han Goh - 4 years, 5 months ago

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Yes. We are performing the bisection method on loga<nlog7<logb \log a < n \log 7 < \log b . The midpoint of this interval is loga+logb2=logab \frac{ \log a + \log b } { 2 } = \log \sqrt{ ab } . We then compare this to nlog7 n \log 7 , which is in essence comparing ab ab to 72n 7^{2n} .

In this way, we don't have to think about what the next number is, or figure out a fancy combination to bound it tightly. We just have to do the bisection method many many times (or automate it with a program), to get the interval down to as small as we want it to.

Calvin Lin Staff - 4 years, 5 months ago

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@Calvin Lin "It is independent of the number of sig figs / decimal places that we started off with."

I was under the impression, you can only find up to the number of SF given. But this just blew my mind. It's like saying:

"Hey, my calculator is kinda bad, it only let shows that log2=0.3 \log 2 = 0.3 and log13=1.1 \log 13 = 1.1 to one decimal place, can you help me find the value of log11 \log 11 to a gazillion decimal places?"

Thanks, I learned something new today!!

Pi Han Goh - 4 years, 5 months ago

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@Pi Han Goh Indeed. See my conversation with Kenny Lau below. It is a common misconception (likely from Physics error analysis) that your calculations cannot be more accurate than the degree of significance that you started out with.

For example, even if your calculator can only multiply whole numbers, and you want to approximate 2 \sqrt{2} , you can do much better than 1<2<2 1 < \sqrt{2} < 2 . For example, by showing that 14142<2000000<14152 1414^2 < 2000000 < 1415^2 , we can conclude that 1.414<2<1.415 1.414 < \sqrt{2} < 1.415 . We can get 2 \sqrt{2} to any degree of accuracy (with sufficient patience / computing power).

Calvin Lin Staff - 4 years, 5 months ago

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@Calvin Lin 543543 543543

Pi Han Goh - 4 years, 5 months ago

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Well, any method based on the two given information can only be correct to 3 significant figures (or decimal places), because the information itself is correct to only 3 significant figures.

Kenny Lau - 4 years, 5 months ago

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Not necessarily. For example, if we know log710 \log 7 ^ {10} to 3 decimal places, then we can get log7 \log 7 to 4 decimal places.

Calvin Lin Staff - 4 years, 5 months ago

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It would not be accurate to 3 decimal places. Log 7^10 would only be accurate to 3 significant figures which is 2 decimal places.

Kenny Lau - 4 years, 5 months ago

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@Kenny Lau I've added a way to show how we can get an arbitrarily tight bound for log7 \log 7 . It is independent of the number of sig figs / decimal places that we started off with.

Calvin Lin Staff - 4 years, 5 months ago

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