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Difference between revisions of "Logarithm"

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== 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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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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== Videos On Logarithms==
  
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==Logarithmic Properties==
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[https://youtu.be/fU7B2H4JJCU Introduction to 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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[https://youtu.be/tUUynpSD5DM Logarithms Properties]
  
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<center><math> \log_a{b} + \log_a{c} = \log_a{bc}.</math></center>
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==Powerful use of logarithms==
  
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Using the laws of exponents, we can derive and prove the following identities:
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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
 +
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)). ex.
  
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*<math>\log_a b^n=n\log_a b</math>
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<cmath>\log_4(3)\approx 0.7924812503605780907268694720\ldots</cmath>
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*<math>\log_a b+ \log_a c=\log_a bc</math>
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<cmath>\log_4(5)\approx 1.160964047443681173935159715\ldots</cmath>
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*<math>\log_a b-\log_a c=\log_a \frac{b}{c}</math>
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<cmath>\log_4(6)\approx 1.292481250360578090726869472\ldots</cmath>
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*<math>(\log_a b)(\log_c d)= (\log_a d)(\log_c b)</math>
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<cmath>\log_4(\log_4(3))\approx -0.1677756462730553083259853611\ldots</cmath>
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*<math>\frac{\log_a b}{\log_a c}=\log_c 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 exercises.
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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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<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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 +
 
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===Discrete Logarithm===
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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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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>.
  
 
== 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>.
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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>.
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== 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 <math>e</math> and the symbol <math>\ln</math> is not used at all.  (This is an example of conflicting [[mathematical convention]]s.)
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<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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# Evaluate <math>(\log_2 3)(\log_3 4)(\log_4 5)\cdots(\log_{2005} 2006)</math>.
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All logarithms are undefined in nonpositive reals, as they are complex. From the identity <math>e^{i\pi}=-1</math>, we have <math>\ln (-1)=i\pi</math>. Additionally, <math>\ln (-n)=\ln n+i\pi</math> for positive real <math>n</math>.
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== Problems ==
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=== 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]]
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=== 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>
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[[2006_AIME_I_Problems/Problem_9 | Source]]
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* The solutions to the system of equations
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<math>\log_{225}x+\log_{64}y=4</math>
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<math>\log_{x}225-\log_{y}64=1</math>
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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>.
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[[2002_AIME_I_Problems/Problem_6 | Source]]
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=== Olympiad ===
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==Video Explanation==
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Five-minute Intro to Logarithms w/ examples [https://youtu.be/CxXc_moY-3Q]
 +
 
 +
==External Links==
 +
 
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Two-minute Intro to Logarithms [http://www.youtube.com/watch?v=ey7ttABX9SM]
  
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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>.
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[[Category:Definition]]
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[[Category:Functions]]

Latest revision as of 11:54, 16 October 2023


Videos On Logarithms

Introduction to Logarithms

Logarithms Properties

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 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)). 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

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 $(\log_{50}{2.5})(\log_{2.5}e)(\ln{2500})$.
  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

  • 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

Olympiad

Video Explanation

Five-minute Intro to Logarithms w/ examples [1]

External Links

Two-minute Intro to Logarithms [2]

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