Approximating \log_{10}7

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!

Markdown

Appears as

*italics* or _italics_

italics

**bold** or __bold__

bold


- bulleted
- list

bulleted
list


1. numbered
2. list

numbered
list

Note: you must add a full line of space before and after lists for them to show up correctly

paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)

example link

> This is a quote

This is a quote

    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"

# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"

Math

Appears as

Remember to wrap math in $ ... $ or \[ ... \] to ensure proper formatting.

2 \times 3

2 \times 3

2^{34}

2^{34}

a_{i-1}

a_{i-1}

\frac{2}{3}

\frac{2}{3}

\sqrt{2}

\sqrt{2}

\sum_{i=1}^3

\sum_{i=1}^3

\sin \theta

\sin \theta

\boxed{123}

\boxed{123}

Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

Markdown	Appears as
`italics` or `_italics_`	italics
`bold` or `__bold__`	bold
- bulleted - list	bulleted list
1. numbered 2. list	numbered list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1 paragraph 2	paragraph 1 paragraph 2
`[example link](https://brilliant.org)`	example link
`> This is a quote`	This is a quote
# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"	# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"

Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

Sort by:

Top Newest

$\begin{aligned} 2400 & \approx & 2401 \\ \log 2400 & \approx & \log 2401 \\ \log (2^3 \times 3 \times 10^2) & \approx & \log 7^4 \\ \log 2^3 + \log 3 + \log 10^2 & \approx & 4 \log 7 \\ 3 \log 2 + \log 3 + 2 & \approx & 4 \log 7 \\ \log7 & \approx & 0.8450 \\ \end{aligned}$

Pi Han Goh - 4 years, 8 months ago

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.

Calvin Lin Staff - 4 years, 8 months ago

I just found numbers that are close to powers of $7$ that are can be factored in terms of $2,3,5,7$
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. =)

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

A Former Brilliant Member - 4 years, 8 months ago

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

Trevor B. - 4 years, 8 months ago

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

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 $49^2$ .

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 $7^{2n}$

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.

@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 $\log 2 = 0.3$ 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!!

@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 $\sqrt{2}$ , 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).

@Calvin Lin – 543543 543543

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, 8 months ago

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

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

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Approximating log⁡107\log_{10}7log10​7

Comments

Approximating $\log_{10}7$