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Algebra

Logarithms- basics

Author: oLahav

The Logarithms

Introducing- logs

Logarithms, which we'll refer to as logs, are an important mathematical concept. Logs are very simply a way to represent a mathematical relation. They're a lot loss scary than you may think. No reason to freak out.

Let's go over basic stuff- how can I represent the number 4? Well, there are infinitely many ways to do so. 1+3

is one way. 2 * 2

is another. So is $\frac{12}{3}$ , and 2 ^2

. And so is $\log_{2} 16$ .

Oh no! Freak out! Ahh!

No, that's a joke. See, logs are just one more way to say 4, or 20, or other numbers. In that case, you may think logs are useless. Well, let's look at the formal definition:

If x = b ^y

, then $y=\log_b x$ .So for example, we all know that 16=2 ^4

. Thus, $4=\log_2 16$ . See, it really is simple! So $\log_5 125=3$ and $\log_4 4096=6$ .

Logs are super useful when you want to know something like this- for what x will 2 ^x = 600

? That's simple, for $x=\log_2 600$ , which you can find out using a scientific calculator. The answer will be something like 9.22. So you see, logs are really useful in some situations.

Log rules! (literally)

There are a few important log rules you should know:

1. $\log_a a ^x = x$ , which makes sense because a ^x=a ^x .

2. $\log_a (x * y)=\log_a x+\log_a y$ . This is because $a ^x * a ^y = a ^{(x+y)}$ .

3. $\log_a \frac{x}{y} = \log_a x - \log_a y$ .

4. $\log_a x ^n= n\log_a x$ (most useful rule!)

5. $\log_a x=\frac{\log_b x}{\log_b a}$ (change of base formula).

The change of base formula is extra-useful because calculators usually give only $\log_{10}$ results (generally speaking, log with no base specified is taken as log base 10).

Also, note the domain of the inside of the log. If we have $\log_a x$ , then x > 0

. This makes intuitive sense looking at the basic definition of the log.

Let's do some practice!

Ex: Solve for x, if $\log_7 (x ^3 + 27) - \log_7 (x+3) = 2$ .

Super easy! Let's first use rule 3. We have: $\log_7 \frac{x ^3 + 27}{x+3}=2$ . Notice that by cube laws, x ^3+27=(x+3)(x ^2 -3x+9) , so we're left with $\log_7 (x ^2-3x+9)=2$ .