Difference between revisions of "Logarithm"

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A '''logarithm''' is a shorthand way of expressing [[exponentional notation]].  
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== Introduction ==
== Introductory ==
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'''Logarithms''' and [[exponents]] are very closely related.  In fact, they are [[inverse functions]]. Basically, this means that logarithms can be used to reverse the result of exponentiation and vice versa just as addition can be used to reverse the result of subtraction.  Thus, if we have <math> a^x = b </math>, then taking the logarithm with base <math> a</math> on both sides will give us
The general form for a logarithm can be expressed as <math>\log_x y=z</math> which means <math>x^z=y</math>. We would read this as "The logarithm of y base x is z".
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We have <math>3^3=27</math>. To express this in [[Logarithmic notation]], we would write it as <math>\log_3 27=3</math>.
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<center><math>\displaystyle x=\log_a{b}.</math></center>
When a logarithm has no base, it is assumed to be base 10.  
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We would read this as "the logarithm of b base a is x". For example, we know that <math>3^3=27</math>. To express this in [[Logarithmic notation]], we would write it as <math>\log_3 27=3</math>.
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When a logarithm has no base, it is assumed to be base 10. Thus, <math>\log(100)</math> means <math>\log_{10}(100)=2</math>.
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==Logarithmic Properties==
 
==Logarithmic Properties==
These hold for all logarithms.
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We can use the properties of exponents to build a set of properties for logarithms.
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We know that <math>a^x\cdot a^y=a^{x+y}</math>.  We let <math> a^x=b</math> and <math> a^y=c </math>.  This also makes <math>\displaystyle a^{x+y}=bc </math>.  From <math> a^x = b</math> we have <math> x = \log_a{b}</math> and from <math> a^y=c </math> we have <math> y=\log_a{c} </math>.  So <math> x+y = \log_a{b}+\log_a{c}</math>.  But we also have from <math>\displaystyle a^{x+y} = bc</math> that <math> x+y = \log_a{bc}</math>  Thus, we have found two expressions for <math> x+y</math> establishing the identity:
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<center><math> \log_a{b} + \log_a{c} = \log_a{bc}.</math></center>
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Using the laws of exponents, we can derive and prove the following identities:
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*<math>\log_a b^n=n\log_a b</math>
 
*<math>\log_a b^n=n\log_a b</math>
 
*<math>\log_a b+ \log_a c=\log_a bc</math>
 
*<math>\log_a b+ \log_a c=\log_a bc</math>
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*<math>(\log_a b)(\log_c d)= (\log_a d)(\log_c b)</math>
 
*<math>(\log_a b)(\log_c d)= (\log_a d)(\log_c b)</math>
 
*<math>\frac{\log_a b}{\log_a c}=\log_c b</math>
 
*<math>\frac{\log_a b}{\log_a c}=\log_c b</math>
*<math>\log_a^n b^n=\log_a b</math>
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*<math>\displaystyle \log_{a^n} b^n=\log_a b</math>
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Try proving all of these as excercises.
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== Problems ==
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# Evaluate <math>(\log_{50}{2.5})(\log_{2.5}e)(\ln{2500}).</math>
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# Simplify <math>\displaystyle \frac 1{\log_2 N}+\frac 1{\log_3 N}+\frac 1{\log_4 N}+\cdots+ \frac 1{\log_{100}}{N} </math> where <math> N=(100!)^3</math>.

Revision as of 09:37, 22 June 2006

Introduction

Logarithms and exponents are very closely related. In fact, they are inverse functions. Basically, this means that logarithms can be used to reverse the result of exponentiation and vice versa just as addition can be used to reverse the result of subtraction. Thus, if we have $a^x = b$, then taking the logarithm with base $a$ on both sides will give us

$\displaystyle x=\log_a{b}.$

We would read this as "the logarithm of b base a is x". For example, we know that $3^3=27$. To express this in Logarithmic notation, we would write it as $\log_3 27=3$.

When a logarithm has no base, it is assumed to be base 10. Thus, $\log(100)$ means $\log_{10}(100)=2$.

Logarithmic Properties

We can use the properties of exponents to build a set of properties for logarithms.

We know that $a^x\cdot a^y=a^{x+y}$. We let $a^x=b$ and $a^y=c$. This also makes $\displaystyle a^{x+y}=bc$. From $a^x = b$ we have $x = \log_a{b}$ and from $a^y=c$ we have $y=\log_a{c}$. So $x+y = \log_a{b}+\log_a{c}$. But we also have from $\displaystyle a^{x+y} = bc$ that $x+y = \log_a{bc}$ Thus, we have found two expressions for $x+y$ establishing the identity:

$\log_a{b} + \log_a{c} = \log_a{bc}.$

Using the laws of exponents, we can derive and prove the following identities:

  • $\log_a b^n=n\log_a b$
  • $\log_a b+ \log_a c=\log_a bc$
  • $\log_a b-\log_a c=\log_a \frac{b}{c}$
  • $(\log_a b)(\log_c d)= (\log_a d)(\log_c b)$
  • $\frac{\log_a b}{\log_a c}=\log_c b$
  • $\displaystyle \log_{a^n} b^n=\log_a b$

Try proving all of these as excercises.

Problems

  1. Evaluate $(\log_{50}{2.5})(\log_{2.5}e)(\ln{2500}).$
  2. Simplify $\displaystyle \frac 1{\log_2 N}+\frac 1{\log_3 N}+\frac 1{\log_4 N}+\cdots+ \frac 1{\log_{100}}{N}$ where $N=(100!)^3$.
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