Working with Logarithms, Rules to Let You Master the Power of Exponents

08.05.2018 |

Episode #9 of the course Foundations of mathematics by John Robin


Hello! Today, we’ll continue working with logarithms.


Basic Rules

Let’s open today’s lesson with a question, then spend the lesson trying to answer it. In the process, I’m going to teach you the basic rules of mastering logarithms.

What is log3(81 x 27)?

Let’s look at log381 and log327 separately to see if we can spot a pattern:

• log381 is the exponent we must put on 3 to turn it into 81. 3x3x3x3 = 81, so 34 = 81, i.e., log381 = 4.

• log327 is the exponent we must put on 3 to turn it into 27. 3x3x3 = 27, so 33 = 27, i.e., log327 = 3.

Now, if we multiply 87×27, this is (3x3x3x3)x(3x3x3). There are seven 3s here, so this is 37. So, log3(81×27) = log3(37) = 7, i.e., we must multiply 3 by itself seven times to turn it into 37.

Did you notice that 7 = 4 + 3?

This is because of our first rule:

1. logc(axb) = logca + logcb, where a and b can be any number and c is any base.

For example:

log (50×120) = log 50 + log 120 = 1.69897 … + 2.07918 …= 3.77815 …

You can verify that this is the same as log (50×120) = log (6000). Try this yourself on your calculator.

We can extend these rules further:

2. logc(a/b) = logca – logcb, where again, a and b can be any number and c is any base.

For example:

log8(64/512) = log864 – log8512 = log8(82) – log8(83) = 2 – 3 = -1

You can verify this by trying it in your calculator:

log8(64/512) = log8(0.125)

And here, since our calculator only gives us log base 10, we’ll use our trick from yesterday, i.e., divide log of the big number by the log of the base.

log8(0.125) = (log 0.125 / log 8) = (-0.90308999 / 0.90308999) = -1

(But bonus points if you spotted that 0.125 is 1/8 and so the exponent required to “grow” 8 into 1/8 is -1.)

3. logcab = b logca

We saw this third rule at play above in our example for log3(81×27) = log3(37) = 7. Notice that this is the same as writing 7 x log33, since log33 = 1. Usually, this is more useful if we had something like log4(256256).

256256 is such a big number, your calculator can’t compute it. So, we’d be staring at a dead end here. But since we now know that log4(256256) = 256 x log4256, we have victory in sight: 256 = 44, so this means log4256 = 4, i.e., the exponent on 4 that turns it into 256.

Therefore, log4(256256) = 256 x log4(256) = 256 x 4 = 1024.

Because of our new rule, this problem just went from impossible to easy.



The true power of these rules can be demonstrated when you run into something like this: log5(875/245).

We can quickly break this into two logarithms with Rule 2:

log5875 – log5245

Now we can divide these numbers into 5 to see what prime number products they break down into:

log553(7) – log55(72)

Now we can use Rule 1 to break up the products in each logarithm into sums:

log553 + log57 – ( log55 + log572)

Now we can use Rule 3 to multiply the logarithms with exponents by the exponent:

3xlog55 + log57 – (log55 + 2xlog57)

Now we can simplify, since log55 = 1.

Everything reduces to:

3 + log57 – 1 – 2xlog57

Notice here that we’ve subtracted 2xlog57. This is because we broke log55(72) into two parts, but the whole thing is subtracted from log553(7), so we must take away log55, as well as log572. (This is why I put brackets around it above, so we wouldn’t forget.)

The final step now is to recognize that we can rearrange the terms:

3 – 1 + log57 – 2xlog57 = 2 – log57

This might not look any better, but if you weren’t allowed a calculator and you had to guess what log5(875/245) might be, this is by far a nicer form.

So, that’s it for logarithms! Tomorrow, we will wrap everything up and see where everything we’ve learned so far leads us, particularly, the kingdom of variables and algebra.


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