# 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 log_{3}(81 x 27)?

Let’s look at log_{3}81 and log_{3}27 separately to see if we can spot a pattern:

• log_{3}81 is the exponent we must put on 3 to turn it into 81. 3x3x3x3 = 81, so 3^{4} = 81, i.e., log_{3}81 = 4.

• log_{3}27 is the exponent we must put on 3 to turn it into 27. 3x3x3 = 27, so 3^{3} = 27, i.e., log_{3}27 = 3.

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

Did you notice that 7 = 4 + 3?

This is because of our first rule:

**1. log _{c}(axb) = log_{c}a + log_{c}b**, 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_{c}a – log_{c}b**, where again,

**a**and

**b**can be any number and

**c**is any base.

For example:

log_{8}(64/512) = log_{8}64 – log_{8}512 = log_{8}(8^{2}) – log_{8}(8^{3}) = 2 – 3 = -1

You can verify this by trying it in your calculator:

log_{8}(64/512) = log_{8}(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.

log_{8}(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 _{c}a^{b} = b log_{c}a**

We saw this third rule at play above in our example for log_{3}(81×27) = log_{3}(3^{7}) = 7. Notice that this is the same as writing 7 x log_{3}3, since log_{3}3 = 1. Usually, this is more useful if we had something like log_{4}(256^{256}).

256^{256} 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 log_{4}(256^{256}) = 256 x log_{4}256, we have victory in sight: 256 = 4^{4}, so this means log_{4}256 = 4, i.e., the exponent on 4 that turns it into 256.

Therefore, log_{4}(256^{256}) = 256 x log_{4}(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: log_{5}(875/245).

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

log_{5}875 – log_{5}245

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

log_{5}5^{3}(7) – log_{5}5(7^{2})

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

log_{5}5^{3} + log_{5}7 – ( log_{5}5 + log_{5}7^{2})

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

3xlog_{5}5 + log_{5}7 – (log_{5}5 + 2xlog_{5}7)

Now we can simplify, since log_{5}5 = 1.

Everything reduces to:

3 + log_{5}7 – 1 – 2xlog_{5}7

Notice here that we’ve subtracted 2xlog_{5}7. This is because we broke log_{5}5(7^{2}) into two parts, but the whole thing is subtracted from log_{5}5^{3}(7), so we must take away log_{5}5, as well as log_{5}7^{2}. (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 + log_{5}7 – 2xlog_{5}7 = 2 – log_{5}7

This might not look any better, but if you weren’t allowed a calculator and you had to guess what log_{5}(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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