Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "Logarithm"

(→‎Problems)
Line 26: Line 26:
 
*<math>\log_{a^n} b^n=\log_a b</math>
 
*<math>\log_{a^n} b^n=\log_a b</math>
 
*<math>\log_{1/a} b=-\log_a b</math>
 
*<math>\log_{1/a} b=-\log_a b</math>
 +
 +
==Powerful use of logarithms==
 +
 +
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 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.
 +
 +
<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>
 +
 +
Therefore by : <cmath>(a^b)^c=a^{bc}</cmath> and identities 1 and 2 above ( 2 being used twice) we get:
 +
 +
<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>
  
 
== Problems ==
 
== Problems ==

Revision as of 13:58, 5 March 2020

Logarithms and exponents are very closely related. In fact, they are inverse functions. 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 $x=\log_a{b}$.

We would read this as "the logarithm of b, base a, is x". For example, we know that $3^4=81$. To express the same fact in logarithmic notation we would write $\log_3 81=4$.


Conventions

Depending on the field, the symbol $\log$ without a base can have different meanings. Sometimes in high schools, the symbol is used to refer to a base 10 logarithm. Thus, $\log(100)$ can mean $\log_{10}(100)=2$. In these contexts, the symbol $\ln$ (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 $\log$ to mean the natural logarithm and do not use the symbol $\ln$. (This is an example of conflicting mathematical conventions.) In addition, the notation $\lg$ is often used by combinatorists and computer scientists to refer to the logarithm base $2$. Occasionally, the base of the logarithms is irrelevant.

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 $a^{x+y}=bc$. 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 $a^{x+y} = bc$ 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, one can derive and prove the following identities:

These formulas also have a number of common special cases:

  • $\log_{a}b=\frac{1}{\log_{b}a}$ (sometimes known as the inverse property of logarithms)
  • $\log_{a^n} b^n=\log_a b$
  • $\log_{1/a} b=-\log_a b$

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 :\[(^24)^x=4^{4^x}\approx(3^5)^6\]

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]

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=Logarithm&oldid=118800"