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List of logarithmic identities
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== 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 β 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.
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