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

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

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

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

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

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

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TopNewest\[\begin{eqnarray} 2400 & \approx & 2401 \\ \log 2400 & \approx & \log 2401 \\ \log (2^3 \times 3 \times 10^2) & \approx & \log 7^4 \\ \log 2^3 + \log 3 + \log 10^2 & \approx & 4 \log 7 \\ 3 \log 2 + \log 3 + 2 & \approx & 4 \log 7 \\ \log7 & \approx & 0.8450 \\ \end{eqnarray} \]

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So, that begs the question of

1. How do I make such an amazing observation that \( 2400 \approx 2401 \)?

2. How do I know what the next approximation should be?

Think about this some, before reading my other comment.

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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: \(6^5 \approx 7777 = \frac {7}{9}(10^4 - 1) \approx \frac {7}{9} \times 10^4 \) or \(6^7 \approx 280000 = 10^4 \times 2^2 \times 7 \), but both methods yields \(0.844\) as the approximation, so I'm kinda stumped right now.

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

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

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

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

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

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

Explicitly, we can use the following series of inequalities:

Step 1: \( 6 < 7 < 8 \). Comparing \( 6 \times 8 \) with \( 7^2 \), we have \( 48 < 7^2 \). Hence, this leads to:

Step 2: \( 48 < 7^2 < 8^2 = 64\). Comparing \( 48 \times 64 \) with \( 7^4 \), we have \( 7^4 < 3072 \). Hence, this leads to:

Step 3: \( 2304 = 48^2 < 7^4 < 3072 \). Comparing \( 2304 \times 3072 \) with \( 7 ^ 8 \), we have \( 7^ 8 < 7077888 \). Hence this leads to:

Step 4: \( 5308416 = 2304^2 < 7^8 < 7077888 \).

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

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 \( 48 < 49 < 50 \), and so we should compare \( 48 \times 50 \) with \( 49^2 \).

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

Taking log to both sides \( \log a < n \log 7 < \log b \)

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

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

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

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.

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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 \( \log 2 = 0.3 \) and \( \log 13 = 1.1 \) to one decimal place, can you help me find the value of \( \log 11 \) to a gazillion decimal places?"

Thanks, I learned something new today!!

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

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543543

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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.

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

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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.

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