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

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. log_c(axb) = log_ca + log_cb, 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. log_c(a/b) = log_ca – log_cb, 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. log_ca^b = b log_ca

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.

 

Example

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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