Introduction

Welcome to Topic 2.12. In this topic, we will learn how to determine and justify the domain of expressions that involve logarithms. Due to logarithmic functions only being defined for positive inputs, logs cannot work with negative values.

Fundamental Rules of Logs

For $x>0,y>0,b>0,b\ne 1$:

Product rule

${\log }_b\left(xy\right)={\log }_b\left(x\right)+{\log }_b\left(y\right)$

Derivation:

Let’s set ${\log }_b(x)=m$ and ${\log }_b(y)=n$. Now let’s have the exponential identity of $b^m\cdot b^n=xy$.

We know this is true since if we substitute $m$ and $n$, we get $b^{{\log }_b(x)}\cdot b^{lo{g}_b(y)}=xy$.

For $b^m\cdot b^n=xy$, let’s change it to $b^{m+n}=xy$ according to exponent rules and then take the logarithm of base $b$ on both sides to get ${\log }_b{b}^{m+n}={\log }_{b}xy\to m+n={\log }_{b}xy$.

Finally, substitute back in $m$ and $n$ to get ${\log }_b(x)+{\log }_b(y)={\log }_{b}xy$

Pretty rad.

Quotient rule

${\log }_b\left(\frac{x}{y}\right)={\log }_b\left(x\right)-{\log }_b\left(y\right)$

Derivation:

Same concept, set ${\log }_b(x)=m$ and ${\log }_b(x)=n$.

Use the exponential identity of $b^{m-n}=\frac{x}{y}$ and this works for the same reason of substituting $m$ and $n$ into the equation.

Take the log of base $b$ on both sides to get ${\log }_b{b}^{m-n}={\log }_b\frac{x}{y}\to m-n={\log }_b\frac{x}{y}$.

Substitute $m$ and $n$ to get ${\log }_b(x)-{\log }_b(y)={\log }_b\frac{x}{y}$

Power rule

${\log }_b\left(x^n\right)=n{\log }_b\left(x\right)$

Derivation:

We know from a previous rule that ${\log }_b(x)+{\log }_b(y)={\log }_b(xy)$.

Now, what if $x$ and $y$ were the same? That means ${\log }_b(x)+{\log }_b(x)={\log }_b(x^2)\to 2{\log }_b(x)={\log }_b(x^2)$.

We can further apply the rule so that there is an $n$ amount of ${\log }_b(x)$ on the left side, so ${\log }_b(x)+{\log }_b(x)+{\log }_b(x)\dots$ a certain amount of times. On the right side, this translates to ${\log }_b(x\cdot x\cdot x\dots )$ a certain amount of times. This means that $n({\log }_b(x))={\log }_b(x^n)$.

Inverse/Exponential Rule

${\log }_b\left(b^x\right)=x,b^{{\log }_b\left(x\right)}=x$

This derivation will be left as an exercise. This requires somewhat different ideas from the previous derivations.

Anyways, 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 $a,b\ne 1$ and $x>0$

${\log }_b\left(x\right)=\frac{{\log }_a\left(x\right)}{{\log }_a\left(b\right)}$

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,

${\log }_b\left(x\right)=\frac{\ln x}{\ln b}=\frac{{\log }_{10}\left(x\right)}{{\log }_{10}\left(b\right)}=\frac{{\log }_{67}\left(x\right)}{{\log }_{67}\left(b\right)}$

Note, if you ever see $\log$ by itself, it is the same as ${\log }_{10}$. Sometimes, you may also see it as $lg$, although that is rare.

The Natural Logarithm

The natural logarithm is the logarithm with base $e\approx 2.71828$. This comes from the limit expression $\underset{x\to \infty }{\lim }{\left(1+\frac{1}{x}\right)}^x$. It is unnecessary to remember this

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

$\ln \left(x\right)={\log }_{e}x$

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 ${\log }_b\left(A\right)$ 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 ${\log }_b\left(u\right)+{\log }_b\left(v\right)={\log }_b\left(uv\right)$

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 $\log \left(x+2\right)+\log \left(x-4\right)=\log \left(6\right)$ where you are asked to solve for $x$.

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

After you find the separate domains, find the intersection 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.

Quick note:  The symbol, "$\cap$" means “and.” It implies both conditions must be true, making this the most restrictive restriction due to both conditions having to prove true. 

Power rule and even/odd/real exponents

The power rule, ${\log }_b\left(u^n\right)=n{\log }_b\left(u\right)$ 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 $u\ne 0$. 

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

${\log }_b\left(x\right)=\frac{{\log }_a\left(x\right)}{{\log }_a\left(b\right)}$

Here, you must have $x>0,a>0,a\ne 1,b>0,b\ne 1,{\log }_a\left(b\right)\ne 0$.

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.

The idea of watching your domains for logarithms is incredibly important in the next topic for when we begin solving equations, which is why you want to watch out for extraneous solutions.

Practice Problems