Editing List of logarithmic identities (section)

== Using simpler operations ==

Logarithms can be used to make calculations easier.  For example, two numbers can be multiplied just by using a logarithm table and adding. These are often known as logarithmic properties, which are documented in the table below.<ref>{{Cite web|title=4.3 - Properties of Logarithms|url=https://people.richland.edu/james/lecture/m116/logs/properties.html|access-date=2020-08-29|website=people.richland.edu}}</ref> The first three operations below assume that {{math|1=''x'' = ''b''<sup>''c''</sup>}} and/or {{math|1=''y'' = ''b''<sup>''d''</sup>}}, so that {{math|1=log<sub>''b''</sub>(''x'') = ''c''}} and {{math|1=log<sub>''b''</sub>(''y'') = ''d''}}. Derivations also use the log definitions {{math|1=''x'' = ''b''<sup>log<sub>''b''</sub>(''x'')</sup>}} and {{math|1=''x'' = log<sub>''b''</sub>(''b''<sup>''x''</sup>)}}.

:{| cellpadding=3
| <math>\log_b(xy)=\log_b(x)+\log_b(y)</math> || because || <math>b^c b^d=b^{c+d}</math>
|-
| <math>\log_b(\tfrac{x}{y})=\log_b(x)-\log_b(y)</math> || because || <math>\tfrac{b^c}{b^d}=b^{c-d}</math>
|-
| <math>\log_b(x^d)=d\log_b(x)</math> || because || <math>(b^c)^d=b^{cd}</math>
|-
| <math>\log_b\left(\sqrt[y]{x}\right)=\frac{\log_b(x)}{y}</math> || because || <math>\sqrt[y]{x}=x^{1/y}</math>
|-
| <math>x^{\log_b(y)}=y^{\log_b(x)}</math> || because || <math>x^{\log_b(y)}=b^{\log_b(x)\log_b(y)}=(b^{\log_b(y)})^{\log_b(x)}=y^{\log_b(x)}</math>
|-
| <math>c\log_b(x)+d\log_b(y)=\log_b(x^c y^d)</math> || because || <math>\log_b(x^c y^d)=\log_b(x^c)+\log_b(y^d)</math>
|}

Where <math>b</math>, <math>x</math>, and <math>y</math> are positive real numbers and <math>b \ne 1</math>, and <math>c</math> and <math>d</math> are real numbers.

The laws result from canceling exponentials and the appropriate law of indices. Starting with the first law:

:<math>xy = b^{\log_b(x)} b^{\log_b(y)} = b^{\log_b(x) + \log_b(y)} \Rightarrow \log_b(xy) = \log_b(b^{\log_b(x) + \log_b(y)}) = \log_b(x) + \log_b(y)</math>

The law for powers exploits another of the laws of indices:

:<math>x^y = (b^{\log_b(x)})^y = b^{y \log_b(x)} \Rightarrow \log_b(x^y) = y \log_b(x)</math>

The law relating to quotients then follows:

:<math>\log_b \bigg(\frac{x}{y}\bigg) = \log_b(x y^{-1}) = \log_b(x) + \log_b(y^{-1}) = \log_b(x) - \log_b(y)</math>

:<math>\log_b \bigg(\frac{1}{y}\bigg) = \log_b(y^{-1}) = - \log_b(y)</math>

Similarly, the root law is derived by rewriting the root as a reciprocal power:

:<math>\log_b(\sqrt[y]x) = \log_b(x^{\frac{1}{y}}) = \frac{1}{y}\log_b(x)</math>

=== Derivations of product, quotient, and power rules ===

These are the three main logarithm laws/rules/principles,<ref>
{{Cite web
|url=https://courseware.cemc.uwaterloo.ca/8/assignments/210/6
|title=Properties and Laws of Logarithms
|website=courseware.cemc.uwaterloo.ca/8
|access-date=2022-04-23
}}
</ref> from which the other properties listed above can be proven. Each of these logarithm properties correspond to their respective exponent law, and their derivations/proofs will hinge on those facts. There are multiple ways to derive/prove each logarithm law – this is just one possible method.

==== Logarithm of a product ====

To state the ''logarithm of a product'' law formally:

:<math>\forall b \in \mathbb{R}_+, b \neq 1, \forall x, y, \in \mathbb{R}_+, \log_b(xy) = \log_b(x) + \log_b(y)</math>

Derivation:

Let <math>b \in \mathbb{R}_+</math>, where <math>b \neq 1</math>,
and let <math>x, y \in \mathbb{R}_+</math>. We want to relate the expressions <math>\log_b(x)</math> and <math>\log_b(y)</math>.  This can be done more easily by rewriting in terms of exponentials, whose properties we already know. Additionally, since we are going to refer to <math>\log_b(x)</math> and <math>\log_b(y)</math> quite often, we will give them some variable names to make working with them easier: Let <math>m = \log_b(x)</math>, and let <math>n = \log_b(y)</math>.

Rewriting these as exponentials, we see that

:<math>\begin{align}
m &= \log_b(x) \iff b^m = x, \\
n &= \log_b(y) \iff b^n = y.
\end{align}</math>

From here, we can relate <math>b^m</math> (i.e. <math>x</math>) and <math>b^n</math> (i.e. <math>y</math>) using exponent laws as

:<math>xy = (b^m)(b^n) = b^m \cdot b^n = b^{m + n}</math>

To recover the logarithms, we apply <math>\log_b</math> to both sides of the equality.

:<math>\log_b(xy) = \log_b(b^{m + n})</math>

The right side may be simplified using one of the logarithm properties from before: we know that <math>\log_b(b^{m + n}) = m + n</math>, giving

:<math>\log_b(xy) = m + n</math>

We now resubstitute the values for <math>m</math> and <math>n</math> into our equation, so our final expression is only in terms of <math>x</math>, <math>y</math>, and <math>b</math>.

:<math>\log_b(xy) = \log_b(x) + \log_b(y)</math>

This completes the derivation.

==== Logarithm of a quotient ====

To state the ''logarithm of a quotient'' law formally:

:<math>\forall b \in \mathbb{R}_+, b \neq 1, \forall x, y, \in \mathbb{R}_+, \log_b \left( \frac{x}{y} \right) = \log_b(x) - \log_b(y)</math>

Derivation:

Let <math>b \in \mathbb{R}_+</math>, where <math>b \neq 1</math>,
and let <math>x, y \in \mathbb{R}_+</math>.

We want to relate the expressions <math>\log_b(x)</math> and <math>\log_b(y)</math>. This can be done more easily by rewriting in terms of exponentials, whose properties we already know. Additionally, since we are going to refer to <math>\log_b(x)</math> and <math>\log_b(y)</math> quite often, we will give them some variable names to make working with them easier: Let <math>m = \log_b(x)</math>, and let <math>n = \log_b(y)</math>.

Rewriting these as exponentials, we see that:
:<math>\begin{align}
m &= \log_b(x) \iff b^m = x, \\
n &= \log_b(y) \iff b^n = y.
\end{align}</math>

From here, we can relate <math>b^m</math> (i.e. <math>x</math>) and <math>b^n</math> (i.e. <math>y</math>) using exponent laws as

:<math>\frac{x}{y} = \frac{(b^m)}{(b^n)} = \frac{b^m}{b^n} = b^{m - n}</math>

To recover the logarithms, we apply <math>\log_b</math> to both sides of the equality.

:<math>\log_b \left( \frac{x}{y} \right) = \log_b \left( b^{m -n} \right)</math>

The right side may be simplified using one of the logarithm properties from before: we know that <math>\log_b(b^{m - n}) = m - n</math>, giving

:<math>\log_b \left( \frac{x}{y} \right) = m -n</math>

We now resubstitute the values for <math>m</math> and <math>n</math> into our equation, so our final expression is only in terms of <math>x</math>, <math>y</math>, and <math>b</math>.

:<math>\log_b \left( \frac{x}{y} \right) = \log_b(x) - \log_b(y)</math>

This completes the derivation.

==== Logarithm of a power ====

To state the ''logarithm of a power'' law formally:

:<math>\forall b \in \mathbb{R}_+, b \neq 1, \forall x \in \mathbb{R}_+, \forall r \in \mathbb{R},  \log_b(x^r) = r\log_b(x)</math>

Derivation:

Let <math>b \in \mathbb{R}_+</math>, where <math>b \neq 1</math>, let <math>x\in \mathbb{R}_+</math>, and let <math>r \in \mathbb{R}</math>. For this derivation, we want to simplify the expression <math>\log_b(x^r)</math>. To do this, we begin with the simpler expression <math>\log_b(x)</math>. Since we will be using <math>\log_b(x)</math> often, we will define it as a new variable: Let <math>m = \log_b(x)</math>.

To more easily manipulate the expression, we rewrite it as an exponential. By definition, <math>m = \log_b(x)  \iff  b^m = x</math>, so we have

:<math>b^m = x</math>

Similar to the derivations above, we take advantage of another exponent law. In order to have <math>x^r</math> in our final expression, we raise both sides of the equality to the power of <math>r</math>:

:<math>
\begin{align}
(b^m)^r &= (x)^r \\
b^{mr} &= x^r
\end{align}
</math>

where we used the exponent law <math>(b^m)^r = b^{mr}</math>.

To recover the logarithms, we apply <math>\log_b</math> to both sides of the equality.

:<math>\log_b(b^{mr}) = \log_b(x^r)</math>

The left side of the equality can be simplified using a logarithm law, which states that <math>\log_b(b^{mr}) = mr</math>.

:<math>mr = \log_b(x^r)</math>

Substituting in the original value for <math>m</math>, rearranging, and simplifying gives

:<math>
\begin{align}
\left( \log_b(x) \right)r &= \log_b(x^r) \\
r\log_b(x) &= \log_b(x^r) \\
\log_b(x^r) &= r\log_b(x)
\end{align}
</math>

This completes the derivation.