2.4 Logarithms


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2.4 Contents


2.4.1 Definition & Notation

Logarithms are inverse functions of exponents $$b^p=x \therefore \log_b x=p$$
Jump to Cologarithms
Example
$$2^3=8 \therefore \log_2 8=3$$

2.4.2 Base Inverse Property

$$\log_b b^p=p$$
Jump to Antilogarithms of an Equation
Jump to Logarithms Power Rule
Jump to Product & Quotient Rules
Jump to Cologarithms
Proof
$$\log_b x = p \land x=b^p \therefore \log_b b^p=p$$
Example
$$\log_2 8=3 \land 8=2^3 \therefore log_2 2^3=3$$

2.4.3 Antilogarithms of an Equation

Given an equality, an equation may be raised as exponential powers with the same base value $$a=b↔x^a=x^b$$
Proof
Given $x^a=x^b$, take the logarithm of base $x$ $$\log_x x^a=\log_x x^b$$ Apply the base inverse property on each side

2.4.4 Power Inverse Property

$$ b^{\log_b x^p}=x^p, \forall x \isin \R$$
Deductive Logic
Logarithms are the inverse functions of exponents, therefore operating a logarithm in an exponent with the same base cancels both
Example
$$\log_2 2^3=2^{log_2 3}=3$$

2.4.5 Power Rule

$$\log_b x^p=p·\log_b x$$
Jump to Change of Base Rule
Jump to Cologarithms
Proof
Let $a=\log_bx$ so that $x=b^a$, then exponentiate to $p$ $$x^p={(b^a)}^p$$ Distribute the power $$x^p=b^{a·p}$$ Take the logarithm using base $b$ $$\log_b x^p=\log_b b^{a·p}$$ Apply the base inverse property $$\log_b x^p=a·p$$ Substitute $a$ for its original form
Example 1
$$\log_2 64=\log_2 2^6=6·\log_2 2=6$$
Example 2
$$\log_2 x^{-1}=-\log_2 x$$

2.4.6 Product & Quotient Rules

$$\log_b (M·N^{\pm 1})=\log_b M \pm \log_b N$$
Reference Plus or Minus Notation
Jump to Sum & Difference Rules
Jump to Cologarithms
Proof
Let $$ x=\log_b M \land ±y=\log_b N^{±1}$$ So that $$M=b^x \land N^{±1}=b^{±y}$$ Multiply the two equalities $$M·N^{±1}=b^x·b^{±y}$$ Apply the exponent power rule $$M·N^{±1}=b^{x±y}$$ Take the logarithm of base $b$ $$\log_b (M·N^{±1})=\log_b b^{x±y}$$ Apply the base inverse property $$\log_b (M·N^{±1})=x±y$$ Substitute $x$ and $±y$ for their original terms

2.4.7 Sum & Difference Rules

$$\log_b (M \pm N)=\log_b M+\log_b \Big(1 \pm \frac{N}{M} \Big)$$
Reference Plus or Minus Notation
Proof
Given $log_b(M±N)$, multiply $N$ by $M/M$ $$\log_b \Big(M \pm \frac{M}{M}·N \Big)$$ Factor $$\log_b \Big(M· \Big( 1 \pm \frac{N}{M} \Big) \Big)$$ Apply the product rule

2.4.8 Change of Base Rule

$$\log_b x = \frac{\log_a x}{\log_a b},$$ $$x>0\medspace\land\medspace a>0\medspace\land\medspace a≠1$$
Properties from the Change of Base Rule
$$1)\qquad \log_b x = \frac{1}{log_x b}$$ $$2)\quad\thinspace \log_{c^n} x=\frac{\log_c x}{n}$$ $$3)\qquad a^{\log_b x}=x^{\log_b a}$$
Proof
Let $c=\log_bx$ so that $b^c=x$, then take the logarithm of base $a$ $$\log_a b^c=\log_a x$$ Apply the power rule $$c·\log_a b=\log_a x$$ Solve for $c$ and sutstitute it for its original form
Proof of Properties
1) Using the change of base rule, substitute $a$ for $x$

2) Using the change of base rule, substitute $c^n$ for $b$ and $c$ for $a$

3) Take the logarithm $b$ of the given formula and apply the power rule

2.4.9 Cologarithms

$$\log_{1/b} x=-\log_b x=log_b x^{-1}$$
Proof
Let $a=log_{1/b} x$ so that by logarithmic definition and applying the negative exponent $$x=b^{–a}$$ Take the logarithm of base $b$ $$\log_b x= \log_b b^{-a}$$ Apply the base inverse property $$\log_b x=-a$$ Negate $$-\log_b x=a$$ Substitute $a$ for $log_{1/b} x$ $$-\log_b x=log_{1/b} x$$ $–log_bx=log_bx^{–1}$ can be proven with both the power and quotient rules

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