Logarithms Lecture 2: Calculating logs mentally

OK, so now I want to go back to figuring out some more logarithms, at least approximately. By the end of all of this, you should be able to figure out the logarithm of any number, in your head, correct to at least one decimal place. This will also give you some practice with the fundamental rule for logs (see the colored box below).

Logs of numbers 1, 2, 3, … , 10

Recall from the last lecture that we decided that the number 3 represents approximately “half a zero”, since $3\times 3=9\simeq 10$ , and multiplying numbers adds the number of zeroes. The logarithm counts the zeroes, so $\log 3 \simeq 0.5$ . (Actually a little less than $0.5$ , because if 3 was exactly half a zero, then 3 x 3 would be exactly 10.)

We can state this a little more mathematically. I’ve said that multiplying numbers adds their number of zeroes. And “log” counts the number of zeroes. So, for example,

log(100 x 1,000) = log(100) + log(1,000)

says, in words, that the number of zeroes in 100 x 1,000, is the number of zeroes in 100 plus the number of zeroes in 1,000 (so 5 = 2 + 3).

In fact, for any two numbers a and b, the number of zeroes in a x b should be the number of zeroes in a plus the number of zeroes in b:

$\log(a\times b)=\log a + \log b$

This is super-important. This is a mathematical way of saying what I’ve said in words, that logs convert multiplication to addition. I’m going to keep coming back to this rule over and over.

This rule allows me to make my argument above look more mathematical. I am setting a=3 and b=3 in the rule. That is, since

$3\times 3 = 9 \simeq 10$ ,

we have

$\log(3\times 3) = \log(9) \simeq \log(10)$ ,

$\log(3\times 3) \simeq 1$ ,

but $\log(3\times 3)=\log 3 + \log 3$ by the fundamental rule:

$\log(3) + \log(3) \simeq 1$ ,

which means that

$\log 3 \simeq \frac{1}{2}$ .

Let’s try to find log 2 in a similar way. We know that

$2\times 2\times 2 = 8 \simeq 10$ ,

where that “approximately equals” is pretty rough this time. That means

$\log(2\times 2\times 2) \simeq \log(10)=1$ ,

and by our fundamental rule, log(2 x 2 x 2) = log 2 + log 2 + log 2. So we know

$\log 2 + \log 2 + \log 2 \simeq 1$ ,

so log 2 must be roughly $\frac{1}{3}$ . It is not exactly $\frac{1}{3}$ , because if it was, then 2 would be “ $\frac{1}{3}$ of a zero”, and 2 x 2 x 2 would be exactly 10. So log 2 must be a bit less than $\frac{1}{3}$ . We could guess

$\log 2\simeq 0.3$ ,

say.

In fact, we can justify this another way. A little experimentation will show that

2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 = 1024,

which is pretty close to 1000. (That’s ten twos multiplied.) Using the rule as we did before (fill in the details!), we end up with

$\log 2 + \log 2 + \log 2 + \log 2 + \log 2 + \log 2 + \log 2 + \log 2 + \log 2 + \log 2 \simeq \log (1000)$ ,

$10\times\log 2 \simeq 3$ .

This gives us the estimate

$\log 2 \simeq 0.3$ .

(Please fill in the details yourself to make sure you understand!)

So, as we go forward, remember the estimates

$\log 2 \simeq 0.3$

$\log 3 \simeq 0.5$

and remember always the fundamental rule,

$\log(a\times b)=\log a + \log b$

Problem 1: Find an estimate for log 4. (Hint: you already have an estimate for log 2; use the fundamental rule.)

Problem 2:
a. Find an estimate for log 6. (Similar hint.)
b. Find an estimate for log 8.

Problem 3: You can find an estimate for log 5 in two ways. (Please do both! They are both informative in different ways.)
a. Make a guess based on your estimates for log 4 and log 6.
b. Use the fact that 2 x 5 = 10, and use the fundamental rule.

Problem 4: Find an estimate for log 7. (Try to do it two different ways, similar to the previous question.)

Problem 5: Now find an estimate for log 9. This one is easiest to do like Problem 3a; I don’t know an easy method like Problem 3b for this one (you can look for one if you would like a challenge problem!).

Problem 6: You know log 10 already. (Right?! Check back if you don’t recall.) The only whole number from 1, 2, 3, … , 10 that we are missing is log 1. It’s a little tricky, so I’ll give away the answer: log 1 turns out to be zero. Use the fundamental rule to show that log 1 has got to be zero.

Problem 7: Finally, summarize all your results. Make a table:

Number:	1	2	3	4	5	6	7	8	9	10
Log:		0.3	0.5

Fill in the rest of the blanks!

Estimating log of any number (larger than 1)

We can use what we’ve done to approximately find the log of any number. (For now I’ll concentrate on numbers larger than 1; I’ll get to smaller numbers later.)

For an example, let me pick a number at random: say, 63,538. What is the log of this number? I know that the number is between 10,000 and 100,000, so it “has 4 point something zeroes”. That is, log(63,538) must be between 4 and 5.

To find out more closely where it lies between 4 and 5, rewrite the number 63,538 as

6.3538 x 10,000

which I’ll round off to

6.4 x 10,000.

That factor of 6.4 is telling us how far we are between 4 zeroes and 5 zeroes, in a multiplicative sense:

$\log(63,538)\simeq\log(6.4 \times 10,000)=\log(10,000)+\log(6.4) =4 + \log(6.4).$

So, the log(6.4) is giving us the decimal part of “4 point something”. Now, we have established that $\log 6\simeq 0.8$ , and $\log 7\simeq 0.85$ , so I’m going to guess that $\log(6.4)\simeq 0.82$ . This finally leads to the estimate

$\log(63,538)\simeq 4.82$ .

Let’s check it:

Log(65,538) on a calculator. Not bad. See below on doing a bit better.

Not bad, right? (See Problem 10 below on how to make the estimate a bit better.)

So, to calculate logs mentally, one way to go is to memorize the logs of 1, 2, 3, … , 10, and then use the method above. My own way is to just memorize $\log 2 \simeq 0.3$ and $\log 3 \simeq 0.5$ , and to figure out the rest of 4, 5, …, 9 as I need them, doing it like you did in the previous problems.

Problem 8: Find the log of each of the following numbers approximately, doing the calculations mentally.
a. 560
b. 4160
c. 31,000,000
d. 72,380,000,000
e. 13,468
Check your answers with a calculator! (On some calculators, the “log” we are using is written “ $\log_{10}$ .”)

Problem 9: Make up some more numbers at random, and calculate the logs of them. This is a good exercise to do while doing something else. You can also do this with a partner: one person makes up a number, the other person works out the log. Note that the numbers can be enormous if you like–that doesn’t make it harder to find the logs. Practice this until you can do it fairly easily.

Problem 10 (optional): You saw in the example above, where I tried to calculate log(63,538), that I was a bit off. The main reason for this is log 3. Our estimate of log 2 was 0.3; the actual value (to four decimal places) is log 2 = 0.3010, so 0.3 is pretty good. But we estimated log 3 = 0.5, and that is a bit off; the actual value is log 3 = 0.4771.

If you want to be accurate to about two decimal places, you can remember $\log 2 \simeq 0.30$ , as above, but remember $\log 3 \simeq 0.48$ , and figure out the other values from these.

Use those two values to recalculate logs of 4, 5, 6, 7, 8, and 9, and make a more accurate version of the table from Problem 7. Use those numbers to re-do the example of log(63,538) that I did above, and see how much closer you get.

(If you want to do even better, you can memorize $\log 2\simeq 0.301$ and $\log3\simeq 0.477$ …)

Logs of numbers 1, 2, 3, … , 10

Estimating log of any number (larger than 1)

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