# 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!

Note by Trevor B.
5 years, 1 month ago

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\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}

- 5 years, 1 month 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?

Staff - 5 years, 1 month ago

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

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: $6^5 \approx 7777 = \frac {7}{9}(10^4 - 1) \approx \frac {7}{9} \times 10^4$ 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. =)

- 5 years, 1 month ago

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

- 5 years, 1 month ago

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

- 5 years, 1 month ago

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

- 5 years, 1 month ago

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 $49^2$.

Staff - 5 years, 1 month ago

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 $7^{2n}$

- 5 years, 1 month ago

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.

Staff - 5 years, 1 month ago

"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!!

- 5 years, 1 month ago

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).

Staff - 5 years, 1 month ago

543543

- 5 years, 1 month ago

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.

- 5 years, 1 month 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.

Staff - 5 years, 1 month ago

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.

- 5 years, 1 month ago

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.

Staff - 5 years, 1 month ago