Difference between revisions of "Logarithm"

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(Logarithmic Properties)
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'''Logarithms''' and [[exponents]] are very closely related.  In fact, they are [[Function/Introduction#The_Inverse_of_a_Function|inverse]] [[function]]s.  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>x=\log_a{b}</math>.
 
  
We would read this as "the logarithm of b, base a, is x".  For example, we know that <math>3^4=81</math>. To express the same fact in logarithmic notation we would write <math>\log_3 81=4</math>.
 
  
  
==Conventions==
 
Depending on the field, the symbol <math>\log</math> without a base can have different meanings.  Typically, in mathematics through the level of [[calculus]], the symbol is used to refer to a base 10 logarithm.  Thus, <math>\log(100)</math> means <math>\log_{10}(100)=2</math>.  Usually, the symbol <math>\ln</math> (an abbreviation of the French "logarithme normal," meaning "natural logarithm") is introduced to refer to the logarithm base [[e]].  However, in higher mathematics such as [[complex analysis]], the base 10 logarithm is typically disposed with entirely, the symbol <math>\log</math> is taken to mean the logarithm base e and the symbol <math>\ln</math> is not used at all.  (This is an example of conflicting [[mathematical convention]]s.)
 
  
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==Powerful use of logarithms==
  
==Logarithmic Properties==
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Some of the real powerful uses of logarithms, come down to never having to deal with massive numbers. ex. :<cmath>((((((3^5)^6)^7)^8)^9)^{10})^{11}=\underbrace{1177\ldots 1}_{\text{793549 digits}}</cmath> would be a pain to have to calculate any time you wanted to use it (say in a comparison of large numbers). its natural logarithm though (partly due to
We can use the properties of exponents to build a set of properties for logarithms.
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left to right parenthesized exponentiation) is only 7 digits before the decimal point. Comparing the logs of the numbers to a given precision  can allow easier comparision than computing and comparing the numbers themselves. Logs also allow (with repetition) to turn left to right exponentiation into power towers (especially useful for tetration (exponentiation repetition with the same exponent)). ex.
  
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>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>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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<cmath>log_4(3)\approx 0.7924812503605780907268694720\ldots</cmath>
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<cmath>log_4(5)\approx 1.160964047443681173935159715\ldots</cmath>
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<cmath>log_4(6)\approx 1.292481250360578090726869472\ldots</cmath>
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<cmath>log_4(log_4(3))\approx -0.1677756462730553083259853611\ldots</cmath>
  
<center><math> \log_a{b} + \log_a{c} = \log_a{bc}.</math></center>
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Therefore by : <cmath>(a^b)^c=a^{bc}</cmath> and identities 1 and 2 above ( 2 being used twice) we get:
  
Using the laws of exponents, we can derive and prove the following identities:
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<cmath>\log_4(\log_4(3))+(\log_4(5)+\log_4(6))\approx 2.285669651531203956336043826\ldots=x</cmath> such that :<cmath>(^24)^x=4^{4^x}\approx(3^5)^6</cmath>
  
*<math>\log_a b^n=n\log_a b</math>
 
*<math>\log_a b+ \log_a c=\log_a bc</math> (Known as the product property.)
 
*<math>\log_a b-\log_a c=\log_a \frac{b}{c}</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>\log_{a^n} b^n=\log_a b</math>
 
  
Try proving all of these as exercises.
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===Discrete Logarithm===
  
Here are some other less useful log properties that follow from these previous ones.
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A only partially related value is the discrete logarithm, used in [[cryptography]] via [[modular arithmetic]]. It's the lowest value <math>c</math> such that, <math>a^c=mx+b</math> for given <math>a,m,b</math> being integers (as well as <math>c,x</math> the unknowns being integer).
*<math>\log_{a}b=\frac{1}{\log_{b}a}</math> (Known as the inverse property)
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*<math>(\log_{a}b)(\log_{b}c)=\log_{a}c</math> (Known as the chain rule.)
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Its related to the usual logarithm, by the fact that if <math>b</math> isn't an integer power of <math>a</math> then <math>\lceil \log_a(m)\rceil</math> is a lower bound on <math>c</math>
*<math>\log_{1/a} b=-\log_a b</math>
 
  
 
== Problems ==
 
== Problems ==
  
 
# Evaluate <math>(\log_{50}{2.5})(\log_{2.5}e)(\ln{2500})</math>.
 
# Evaluate <math>(\log_{50}{2.5})(\log_{2.5}e)(\ln{2500})</math>.
 
 
# Evaluate <math>(\log_2 3)(\log_3 4)(\log_4 5)\cdots(\log_{2005} 2006)</math>.
 
# Evaluate <math>(\log_2 3)(\log_3 4)(\log_4 5)\cdots(\log_{2005} 2006)</math>.
 
 
# Simplify <math>\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>.
 
# Simplify <math>\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>.
 
  
 
== Natural Logarithm ==
 
== Natural Logarithm ==
The natural logarithm of <math>a</math> is <math>\log_e a=\ln a</math>. The function <math>f(x)=\ln x</math> is the inverse of <math>f(x)=e^x</math>.
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The '''natural logarithm''' is the logarithm with base [[e]].  It is usually denoted <math>\ln</math>, an abbreviation of the French ''logarithme normal'', so that <math> \ln(x) = \log_e(x).</math> However, in higher mathematics such as [[complex analysis]], the base 10 logarithm is typically disposed with entirely, the symbol <math>\log</math> is taken to mean the logarithm base e and the symbol <math>\ln</math> is not used at all.  (This is an example of conflicting [[mathematical convention]]s.)
  
 
<math>\ln a</math> can also be defined as the area under the curve <math>y=\frac{1}{x}</math> between 1 and a, or <math>\int^a_1 \frac{1}{x}\, dx</math>.
 
<math>\ln a</math> can also be defined as the area under the curve <math>y=\frac{1}{x}</math> between 1 and a, or <math>\int^a_1 \frac{1}{x}\, dx</math>.
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== Problems ==
 
== Problems ==
 
=== Introductory ===
 
=== Introductory ===
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* What is the value of <math>a</math> for which <math>\frac1{\log_2a}+\frac1{\log_3a}+\frac1{\log_4a}=1</math>?
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[[2015_AMC_12A_Problems/Problem_14 | Source]]
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* Positive integers <math>a</math> and <math>b</math> satisfy the condition <math>\log_2(\log_{2^a}(\log_{2^b}(2^{1000})))=0.</math> Find the sum of all possible values of <math>a+b</math>.
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[[2013_AIME_II_Problems/Problem_2 | Source]]
 
=== Intermediate ===
 
=== Intermediate ===
 
* The [[sequence]] <math> a_1, a_2, \ldots </math> is [[geometric sequence|geometric]] with <math> a_1=a </math> and common [[ratio]] <math> r, </math> where <math> a </math> and <math> r </math> are positive integers. Given that <math> \log_8 a_1+\log_8 a_2+\cdots+\log_8 a_{12} = 2006, </math> find the number of possible ordered pairs <math> (a,r). </math>
 
* The [[sequence]] <math> a_1, a_2, \ldots </math> is [[geometric sequence|geometric]] with <math> a_1=a </math> and common [[ratio]] <math> r, </math> where <math> a </math> and <math> r </math> are positive integers. Given that <math> \log_8 a_1+\log_8 a_2+\cdots+\log_8 a_{12} = 2006, </math> find the number of possible ordered pairs <math> (a,r). </math>
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=== Olympiad ===
 
=== Olympiad ===
  
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==External Links==
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Two-minute Intro to Logarithms [http://www.youtube.com/watch?v=ey7ttABX9SM]
  
 
[[Category:Definition]]
 
[[Category:Definition]]
 
[[Category:Functions]]
 
[[Category:Functions]]

Revision as of 13:01, 17 April 2020



Powerful use of logarithms

Some of the real powerful uses of logarithms, come down to never having to deal with massive numbers. ex. :\[((((((3^5)^6)^7)^8)^9)^{10})^{11}=\underbrace{1177\ldots 1}_{\text{793549 digits}}\] would be a pain to have to calculate any time you wanted to use it (say in a comparison of large numbers). its natural logarithm though (partly due to left to right parenthesized exponentiation) is only 7 digits before the decimal point. Comparing the logs of the numbers to a given precision can allow easier comparision than computing and comparing the numbers themselves. Logs also allow (with repetition) to turn left to right exponentiation into power towers (especially useful for tetration (exponentiation repetition with the same exponent)). ex.

\[log_4(3)\approx 0.7924812503605780907268694720\ldots\] \[log_4(5)\approx 1.160964047443681173935159715\ldots\] \[log_4(6)\approx 1.292481250360578090726869472\ldots\] \[log_4(log_4(3))\approx -0.1677756462730553083259853611\ldots\]

Therefore by : \[(a^b)^c=a^{bc}\] and identities 1 and 2 above ( 2 being used twice) we get:

\[\log_4(\log_4(3))+(\log_4(5)+\log_4(6))\approx 2.285669651531203956336043826\ldots=x\] such that :\[(^24)^x=4^{4^x}\approx(3^5)^6\]


Discrete Logarithm

A only partially related value is the discrete logarithm, used in cryptography via modular arithmetic. It's the lowest value $c$ such that, $a^c=mx+b$ for given $a,m,b$ being integers (as well as $c,x$ the unknowns being integer).

Its related to the usual logarithm, by the fact that if $b$ isn't an integer power of $a$ then $\lceil \log_a(m)\rceil$ is a lower bound on $c$

Problems

  1. Evaluate $(\log_{50}{2.5})(\log_{2.5}e)(\ln{2500})$.
  2. Evaluate $(\log_2 3)(\log_3 4)(\log_4 5)\cdots(\log_{2005} 2006)$.
  3. Simplify $\frac 1{\log_2 N}+\frac 1{\log_3 N}+\frac 1{\log_4 N}+\cdots+ \frac 1{\log_{100}N}$ where $N=(100!)^3$.

Natural Logarithm

The natural logarithm is the logarithm with base e. It is usually denoted $\ln$, an abbreviation of the French logarithme normal, so that $\ln(x) = \log_e(x).$ However, in higher mathematics such as complex analysis, the base 10 logarithm is typically disposed with entirely, the symbol $\log$ is taken to mean the logarithm base e and the symbol $\ln$ is not used at all. (This is an example of conflicting mathematical conventions.)

$\ln a$ can also be defined as the area under the curve $y=\frac{1}{x}$ between 1 and a, or $\int^a_1 \frac{1}{x}\, dx$.

All logarithms are undefined in nonpositive reals, as they are complex. From the identity $e^{i\pi}=-1$, we have $\ln (-1)=i\pi$. Additionally, $\ln (-n)=\ln n+i\pi$ for positive real $n$.

Problems

Introductory

  • What is the value of $a$ for which $\frac1{\log_2a}+\frac1{\log_3a}+\frac1{\log_4a}=1$?

Source

  • Positive integers $a$ and $b$ satisfy the condition $\log_2(\log_{2^a}(\log_{2^b}(2^{1000})))=0.$ Find the sum of all possible values of $a+b$.

Source

Intermediate

  • The sequence $a_1, a_2, \ldots$ is geometric with $a_1=a$ and common ratio $r,$ where $a$ and $r$ are positive integers. Given that $\log_8 a_1+\log_8 a_2+\cdots+\log_8 a_{12} = 2006,$ find the number of possible ordered pairs $(a,r).$

Source

Olympiad

External Links

Two-minute Intro to Logarithms [1]