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Difference between revisions of "Logarithm"
 
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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>.
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'''Logarithms''' are right inverses of powers.
  
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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>.
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== Definition ==
  
 +
Logarithms, the [[right inverse]]s of powers can be written as <math>\log_a(b)</math> for some <math>a</math> and <math>b</math>, where <math>a^{\log_a(b)}=b</math>.
  
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==Conventions==
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== Applications ==
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Depending on the field, the symbol <math>\log</math> without a base can have different meanings.  Sometimes in high schools, the symbol is used to refer to a base 10 logarithm.  Thus, <math>\log(100)</math> can mean <math>\log_{10}(100)=2</math>.  In these contexts, the symbol <math>\ln</math> (an abbreviation of the French "logarithme normal," meaning "natural logarithm") is introduced to refer to the logarithm base [[e]], or '''natural logarithm'''.  However, the choice of base 10 is arbitrary, and convenient only for computations in a base-10 number system.  The natural logarithm, however, has many convenient mathematical properties, so practicing mathematicians often take the symbol <math>\log</math> to mean the natural logarithm and do not use the symbol <math>\ln</math>.  (This is an example of conflicting [[mathematical convention]]s.) In addition, the notation <math>\lg</math> is often used by [[combinatorics | combinatorists]] and computer scientists to refer to the logarithm base <math>2</math>.  Occasionally, the base of the logarithms is irrelevant.
 
  
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==Logarithmic Properties==
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Some of the real powerful uses of logarithms come down to never having to deal with massive numbers. For example,
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We can use the properties of exponents to build a set of properties for logarithms.
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<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
 +
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 comparison 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)). For example,
  
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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>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>
 +
<cmath>\log_4(5)\approx 1.160964047443681173935159715\ldots</cmath>
 +
<cmath>\log_4(6)\approx 1.292481250360578090726869472\ldots</cmath>
 +
<cmath>\log_4(\log_4(3))\approx -0.1677756462730553083259853611\ldots</cmath>
  
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<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:
  
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Using the laws of exponents, one 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>
  
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*<math>\log_a b^n=n\log_a b</math>
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=== Discrete Logarithm ===
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*<math>\log_a b+ \log_a c=\log_a bc</math>
 
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*<math>\log_a b-\log_a c=\log_a \frac{b}{c}</math>
 
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*<math>(\log_a b)(\log_c d)= (\log_a d)(\log_c b)</math>
 
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*<math>\frac{\log_a b}{\log_a c}=\log_c b</math> (the [[change of base formula]])
 
  
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These formulas also have a number of common special cases:
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An 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 integers).
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*<math>\log_{a}b=\frac{1}{\log_{b}a}</math> (sometimes known as the ''inverse property'' of logarithms)
 
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*<math>\log_{a^n} b^n=\log_a b</math>
 
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*<math>\log_{1/a} b=-\log_a b</math>
 
  
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==Powerful use of logarithms==
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It's 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>.
  
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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
+
== Problems ==
−
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.
 
−
 
 
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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>
 
  
−
Therefore by : <cmath>(a^b)^c=a^{bc}</cmath> and identities 1 and 2 above ( 2 being used twice) we get:
+
# Evaluate:
 +
<cmath>
 +
(\log_{50}{2.5})(\log_{2.5}e)(\ln{2500})
 +
</cmath>
  
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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>
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# Evaluate:
 +
<cmath>
 +
(\log_2 3)(\log_3 4)(\log_4 5) \cdots (\log_{2005} 2006)
 +
</cmath>
  
−
== Problems ==
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# Simplify the following expression:
 +
<cmath>
 +
\frac{1}{\log_2 N} + \frac{1}{\log_3 N} + \frac{1}{\log_4 N} + \cdots + \frac{1}{\log_{100} N},
 +
</cmath>
  
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# Evaluate <math>(\log_{50}{2.5})(\log_{2.5}e)(\ln{2500})</math>.
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where \( N = (100!)^3 \).
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# Evaluate <math>(\log_2 3)(\log_3 4)(\log_4 5)\cdots(\log_{2005} 2006)</math>.
 
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# 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 ==
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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.)
+
 
 +
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 <math>e</math> 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>.
Line 55: Line 54:
  
 
== Problems ==
 
== Problems ==
 +
 
=== Introductory ===
 
=== Introductory ===
 +
 
* What is the value of <math>a</math> for which <math>\frac1{\log_2a}+\frac1{\log_3a}+\frac1{\log_4a}=1</math>?
 
* What is the value of <math>a</math> for which <math>\frac1{\log_2a}+\frac1{\log_3a}+\frac1{\log_4a}=1</math>?
 
[[2015_AMC_12A_Problems/Problem_14 | Source]]
 
[[2015_AMC_12A_Problems/Problem_14 | Source]]
 
* 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>.  
 
* 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>.  
−
[[2013_AIME_II_Problems/Problem_2 | Source]]
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[[2013_AIME_II_Problems/Problem_2|Source]]
 +
 
 
=== Intermediate ===
 
=== Intermediate ===
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* 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>
 
−
[[2006_AIME_I_Problems/Problem_9 | Source]]
 
−
=== Olympiad ===
 
  
−
==External Links==
+
* 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>
−
Two-minute Intro to Logarithms [http://www.youtube.com/watch?v=ey7ttABX9SM]
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[[2006_AIME_I_Problems/Problem_9|Source]]
 +
 
 +
* The solutions to the system of equations
 +
<math>\log_{225}x+\log_{64}y=4</math>
 +
 
 +
<math>\log_{x}225-\log_{y}64=1</math>
 +
are <math>(x_1,y_1)</math> and <math>(x_2,y_2)</math>. Find <math>\log_{30}\left(x_1y_1x_2y_2\right)</math>.
 +
[[2002_AIME_I_Problems/Problem_6|Source]]
 +
 
 +
<!------=== Olympiad ===
 +
 
 +
{{problem}}------>
 +
 
 +
== Videos ==
 +
 
 +
* [//youtu.be/CxXc_moY-3Q Five-minute Intro to Logarithms]
 +
* [//youtu.be/fU7B2H4JJCU Introduction to Logarithms]
 +
* [//youtu.be/tUUynpSD5DM Logarithms Properties]
 +
* [//www.youtube.com/watch?v=Z5myJ8dg_rM Khan Academy]
 +
* [//www.youtube.com/watch?v=ey7ttABX9SM Two-minute Intro to Logarithms]
  
 
[[Category:Definition]]
 
[[Category:Definition]]
 
[[Category:Functions]]
 
[[Category:Functions]]
 +
{{stub}}

Latest revision as of 13:47, 17 March 2025

Logarithms are right inverses of powers.

Definition

Logarithms, the right inverses of powers can be written as $\log_a(b)$ for some $a$ and $b$, where $a^{\log_a(b)}=b$.

Applications

Some of the real powerful uses of logarithms come down to never having to deal with massive numbers. For example, \[((((((3^5)^6)^7)^8)^9)^{10})^{11}=\underbrace{1177\ldots 1}_{\text{793549 digits}}\] 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 comparison 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)). For example,

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

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

An 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 integers).

It's 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:

\[(\log_{50}{2.5})(\log_{2.5}e)(\ln{2500})\]

  1. Evaluate:

\[(\log_2 3)(\log_3 4)(\log_4 5) \cdots (\log_{2005} 2006)\]

  1. Simplify the following expression:

\[\frac{1}{\log_2 N} + \frac{1}{\log_3 N} + \frac{1}{\log_4 N} + \cdots + \frac{1}{\log_{100} N},\]

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

  • The solutions to the system of equations

$\log_{225}x+\log_{64}y=4$

$\log_{x}225-\log_{y}64=1$ are $(x_1,y_1)$ and $(x_2,y_2)$. Find $\log_{30}\left(x_1y_1x_2y_2\right)$. Source


Videos

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