Introduction

Welcome to Topic 2.12. In this topic, you will learn how to determine and justify the domain of expressions that involve logarithms. Because logarithmic functions are only defined for positive inputs, any time you work with logs

Fundamental rules of logs

For $x > 0, y > 0, b > 0, b \neq = 1$ :

Product rule

$lo g_{b} (x y) = lo g_{b} (x) + lo g_{b} (y)$

Quotient rule

$lo g_{b} (\frac{x}{y}) = lo g_{b} (x) - lo g_{b} (y)$

Power rule

$lo g_{b} (x^{n}) = n lo g_{b} (x)$

Inverse/exponential rule

$lo g_{b} (b^{x}) = x, b^{l o g_{b} (x)} = x$

These formulas are used to expand or condense logarithmic expressions. Always ensure the arguments of all logarithms remain positive when applying these rules.

Change of Base Formula

For any positive bases $a, b \neq = 1 an d x > 0$

$lo g_{b} (x) = \frac{lo g _{a} ( x )}{lo g _{a} ( b )}$

Depending on your calculator, you might not have the ability to use different bases, rather, only base $e$ and $10$ , and you can use this formula in order to use those custom bases. Thus,

$lo g_{b} (x) = \frac{ln x}{ln b} = \frac{lo g _{10} ( x )}{lo g _{10} ( b )} = \frac{lo g _{67} ( x )}{lo g _{67} ( b )}$

Note, if you ever see $lo g$ by itself, it is the same as $lo g_{10}$ . Sometimes, you may also see it as $l g$ , although that is rare.

The Natural Logarithm

The natural logarithm is the logarithm with base $e \approx 2.71828$ .

It is useful, but not necessary, to memorize the digits of $e$ at least up to two decimal places.

$ln (x) = lo g_{e} x$

Domain Considerations

When you manipulate logarithmic expressions (expand, condense, solve, substitute, or transform graphs), you must track the domain. Remember, the algebraic rules provided for logs above are only valid when the argument is positive.

Basic Principle

A logarithm $lo g_{b} (A)$ exists if and only if its argument $A$ satisfies $A > 0$

So whenever you produce a new logarithmic expression or use a rule, you must keep in mind your domain.

When rules change the apparent domain (and what to do)

Product/Quotient rules

Say you have $lo g_{b} (u) + lo g_{b} (v) = lo g_{b} (uv)$

Then you must require both $u > 0$ and $v > 0$ . After condensing, the requirement becomes $uv > 0$ . However, those are not equivalent logically. $uv > 0$ allows the possibility that $u < 0$ and $v < 0$ , which would make the original two logs undefined. Therefore:

Before you do any altercations or perform any rules, find the domain of your original problem. For example, say you have the problem $lo g (x + 2) + lo g (x - 4) = lo g (6)$ where you are asked to solve for $x$ .

Before you do anything, find the domain of each logarithmic term. The first term as the domain restriction that $x > - 2$ . The second has the restriction $x > 4$ . The third has no $x$ , so it has no restriction since it is just a constant value.

After you find the separate domains, find the intersection ( $\cap$ – the symbol means “and;” it implies both conditions must be true) of all domains. In this case, the final domain for our solutions of $x$ must fall in the range of $x > 4$ . If a solution we find doesn’t satisfy that domain, then it is an extraneous solution and is not an answer.

Power rule and even/odd/real exponents

The power rule, $lo g_{b} (u^{n}) = n lo g_{b} (u)$ requires that $u > 0$ when $n$ is any real number.

If $n$ is an even integer (e.g. 2), then $u^{n} > 0$ for all $u \neq = 0$ .

If $n$ is an odd integer, $u^{n}$ has the same sign as $u$ , so you must still take the domain as you normally would.

If $n$ is rational (e.g. $\frac{1}{2}$ ), $u^{\frac{1}{2}}$ is not real for any negative $u$ . So, simply take the domain how you normally would.

Change of base

$lo g_{b} (x) = \frac{lo g _{a} ( x )}{lo g _{a} ( b )}$

Here, you must have $x > 0, a > 0, a \neq = 1, b > 0, b \neq = 1, lo g_{a} (b) \neq = 0$ .

The last one is implied since it would require $b = 1$ , which is not possible.

When you use change of base, there’s not much to watch out for.

Solving Logarithmic Equations

There are two mistakes people often make when solving Logarithmic equations.

Algebraic manipulations that introduce sign discrepancies. This could be squaring both sides, or condensing strings of logs where the split requires positive factors.
Forgetting domain restrictions. REMEMBER THIS ONE; IT IS THE LARGEST SOURCE OF ERROR.

So, for good practice, before solving any Logarithmic Equation, you should write down all domain inequalities implied by each term. Then, find the intersection ( $\cap$ ) of all the domains. Alternatively, you could also plug all the solutions you find back into the original equation and discard any that make the terms invalid.

Compositions and Substitutions

In some situations, you may have to substitute a variable within a logarithm. In these rare cases, simply remember whatever is inside the logarithm must be positive and you will be fine.

Why do algebraic rules work when the domain changes?

It could be unintuitive to think about. When we manipulate logs, they remain valid, yet their domain changes.

When we apply log rules, we are manipulating expressions that are only defined on certain inputs. The algebraic rules themselves are indeed correct, but only under the domain assumptions that we enforce. If you change the form, you may add extra constraints that were not explicit in the original form.

However, they are still equal.

When you rewrite an expression, you create a new expression that equal the original on the intersection of their domains. There are three things that could happen

Same domain, same values. This means the identity holds everywhere both sides are defined.
The new expression has a smaller domain. The rewritten form is defined on fewer inputs because the algebraic step required extra assumptions, like splitting a logarithmic. The equality is still true for inputs in the new domain (which is contained within the original domain), but the original may have more inputs where the original is defined and the new form is not. So equality holds on the new domain.
The new expression extends domain but equals on overlap. Sometimes you use absolute values to create an identity valid on a wider domain. $lo g x^{2} = 2 lo g ∣ x ∣$ is an example.

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