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Difference between revisions of "User:Temperal/The Problem Solver's Resource2"

(→‎Rules of Exponentiation and Logarithms: explanations)
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===Definitions===
 
===Definitions===
 
*Exponentials: Do you really need this one? If <math>a=\underbrace{b\times b\times b\times \cdots \times b}_{x\text{ }b'\text{s}}</math>, then <math>a=b^x</math>
 
*Exponentials: Do you really need this one? If <math>a=\underbrace{b\times b\times b\times \cdots \times b}_{x\text{ }b'\text{s}}</math>, then <math>a=b^x</math>
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*Logarithms: If <math>b^a=x</math>, <math>\log_b{x}=a</math>. Note that a logarithm in base [[e]], i.e. <math>\log_e{x}=a</math> is denoted as <math>\ln{x}=a</math>, or the natural logarithm of x. If no base is specified, then a logarithm is assumed to be in base 10.
+
*Logarithms: If <math>b^x=a</math>, then <math>\log_b{a}=x</math>. Note that a logarithm in base [[e]], i.e. <math>\log_e{x}=a</math> is denoted as <math>\ln{x}=a</math>, or the natural logarithm of x. If no base is specified, then a logarithm is assumed to be in base 10.
  
 
===Rules of Exponentiation===
 
===Rules of Exponentiation===

Latest revision as of 17:22, 21 January 2016


Introduction | Other Tips and Tricks | Methods of Proof | You are currently viewing page 2.

Exponentials and Logarithms

This is just a quick review of logarithms and exponents; it's elementary content.

Definitions

  • Exponentials: Do you really need this one? If $a=\underbrace{b\times b\times b\times \cdots \times b}_{x\text{ }b'\text{s}}$, then $a=b^x$
  • Logarithms: If $b^x=a$, then $\log_b{a}=x$. Note that a logarithm in base e, i.e. $\log_e{x}=a$ is denoted as $\ln{x}=a$, or the natural logarithm of x. If no base is specified, then a logarithm is assumed to be in base 10.

Rules of Exponentiation

$a^x \cdot a^y=a^{x+y}$

$(a^x)^y=a^{xy}$

$\frac{a^x}{a^y}=a^{x-y}$

$a^0=1$, where $a\ne 0$.

These should all be trivial and easily proven by the reader.

Rules of Logarithms

$\log_b b=1$

This can be seen by writing as $b^1=b$.

$\log_b xy=\log_b x +\log_b y$

$\log_b x^y=y\cdot \log_b x$

$\log_b \frac{x}{y} =\log_b x-\log_b y$

$\log_b a=\frac{1}{\log_a b}$

$\log_b a=\frac{\log_x a}{\log_x b}$, where x is a constant.

All of the above should be proven by the reader without too much difficulty - substitution and putting things in exponential form will help.

$\log_1 a$ and $\log_0 a$ are undefined, as there is no $x$ such that $1^x=a$ except when $a=1$ (in which case there are infinite $x$) and likewise with $0$.

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