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Difference between revisions of "Logarithm"

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== Introduction ==
 
== Introduction ==
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'''Logarithms''' and [[exponents]] are very closely related.  In fact, they are [[inverse function | 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 <math>\displaystyle x=\log_a{b}</math>.
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'''Logarithms''' and [[exponents]] are very closely related.  In fact, they are [[inverse function | 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 <math>\displaystyle x=\log_a{b}</math>.
  
 
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>.
 
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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We can use the properties of exponents to build a set of properties for logarithms.
 
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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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:
  
 
<center><math> \log_a{b} + \log_a{c} = \log_a{bc}.</math></center>
 
<center><math> \log_a{b} + \log_a{c} = \log_a{bc}.</math></center>

Revision as of 11:53, 26 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 exercises.

Problems

  1. Evaluate $(\log_{50}{2.5})(\log_{2.5}e)(\ln{2500})$.
  1. Evaluate $(\log_2 3)(\log_3 4)(\log_4 5)\cdots(\log_{2005} 2006)$.
  1. 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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