Un logaritmo de un número \(x\) es por definición es el exponente al cual hay que elevar un número \(a\) denominado base para obtener dicho número \(x\), es decir, \(\log_{b}{x}={a}\leftrightarrow{b}^{a}={x}\); en donde \(b\in\left(0;1\right)\cup\left(1;\infty\right)\), \(x\in\left(0;\infty\right)\) e \(a\in\left(-\infty;+\infty\right)\)
\[\log_{b}{x}={a}\leftrightarrow{b}^{a}={x}\]
Los logaritmos, independientemente de la base elegida, cumplen una serie de propiedades comunes que los caracterizan:
El logaritmo de su base es siempre \(1\); \(\log_{b}{b}=1\) ya que \({b}^{1}=b\).
El logaritmo de \(1\) es cero \(0\) (independientemente de la base); \(\log_{b}{1}=0\) ya que \({b}^{0}=1\).
El logaritmo de un número negativo no esta definido en los números reales, es decir, los números negativos no tienen logaritmo en \(\mathbf{R}\).
Sean \(b=2\), \(x=8\) cuál será el logaritmo en base \(b\) de \(x\) o el número \(a\) al que se debe elevar \(b\) para obtener \(x\)
\[\log_{2}{8}={3}\leftrightarrow{2}^{3}={8}\]
log(x=8, base=2)## [1] 32*2*2## [1] 82**3## [1] 82^3## [1] 8Sean \(b=2\), \(x=64\) cuál será el logaritmo en base \(b\) de \(x\) o el número \(a\) al que se debe elevar \(b\) para obtener \(x\)
\[\log_{2}{64}={6}\leftrightarrow{2}^{6}={64}\]
log(x=64, base=2)## [1] 62*2*2*2*2*2## [1] 642**6## [1] 642^6## [1] 64Sean \(b=3\), \(x=27\) cuál será el logaritmo en base \(b\) de \(x\) o el número \(a\) al que se debe elevar \(b\) para obtener \(x\)
\[\log_{3}{27}={3}\leftrightarrow{3}^{3}={27}\]
log(x=27, base=3)## [1] 33*3*3## [1] 273**3## [1] 273^3## [1] 27Sean \(b=3\), \(x=81\) cuál será el logaritmo en base \(b\) de \(x\) o el número \(a\) al que se debe elevar \(b\) para obtener \(x\)
\[\log_{3}{243}={5}\leftrightarrow{3}^{5}={243}\]
log(x=243, base=3)## [1] 53*3*3*3*3## [1] 2433**5## [1] 2433^5## [1] 243Sean \(b=5\), \(x=125\) cuál será el logaritmo en base \(b\) de \(x\) o el número \(a\) al que se debe elevar \(b\) para obtener \(x\)
\[\log_{5}{125}={3}\leftrightarrow{5}^{3}={125}\]
log(x=125, base=5)## [1] 35*5*5## [1] 1255**3## [1] 1255^3## [1] 125log(c(5,25,125),base=5)## [1] 1 2 3Sean \(b=5\), \(x=625\) cuál será el logaritmo en base \(b\) de \(x\) o el número \(a\) al que se debe elevar \(b\) para obtener \(x\)
\[\log_{5}{625}={4}\leftrightarrow{5}^{4}={625}\]
log(x=625, base=5)## [1] 45*5*5*5## [1] 6255**4## [1] 6255^4## [1] 625Dentro de los logaritmos se cumplen algunas propiedades útiles a la hora de realizar logaritmos o entender como funcionan en sí mismos. Se destacaran solo algunas:
\[\log_{b}{\left(x \cdot y\right)} = \log_{b}{x}+\log_{b}{y};\ con\ x,y\in\textbf{R}^{+}\ y\ b\in\textbf{R}^{+}\setminus\left\{1\right\}\]
\[\log_{2}{\left(8 \cdot 64\right)} = \log_{2}{8}+\log_{2}{64}=3+6=9\]
log(x=8*64, base=2)## [1] 92**9==(8*64)## [1] TRUElog(x=512, base=2)## [1] 9log(x=8,base=2) + log(x=64,base=2)## [1] 9\[\log_{3}{\left(27 \cdot 243\right)} = \log_{3}{27}+\log_{3}{243}=3+5=8\]
log(x=27*243, base=3)## [1] 83**8==(27*243)## [1] TRUElog(x=6561,base=3)## [1] 8log(x=27,base=3) + log(x=243,base=3)## [1] 8\[\log_{5}{\left(125 \cdot 625\right)} = \log_{5}{125}+\log_{5}{625}=3+4=7\]
log(x=125*625, base=5)## [1] 75**7==(125*625)## [1] TRUElog(x=78125, base=5)## [1] 7log(x=125,base=5) + log(x=625,base=5)## [1] 7\[\log_{b}{\left(\frac{x}{y}\right)} = \log_{b}{x}-\log_{b}{y};\ con\ x,y\in\textbf{R}^{+}\ y\ b\in\textbf{R}^{+}\setminus\left\{1\right\}\]
\[\log_{2}{\left(\frac{64}{8}\right)} = \log_{2}{64}-\log_{2}{8}=6-3=3\]
log(x=64/8, base=2)## [1] 32**3==(64/8)## [1] TRUElog(x=8, base=2)## [1] 3log(x=64,base=2) - log(x=8,base=2)## [1] 3\[\log_{3}{\left(\frac{243}{27}\right)} = \log_{3}{243}-\log_{3}{27}=5-3=2\]
log(x=243/27, base=3)## [1] 23**2==(243/27)## [1] TRUElog(x=9, base=3)## [1] 2log(x=243,base=3) - log(x=27,base=3)## [1] 2\[\log_{5}{\left(\frac{625}{125}\right)} = \log_{5}{625}-\log_{5}{125}=4-3=1\]
log(x=625/125, base=5)## [1] 15**1==(625/125)## [1] TRUElog(x=5, base=5)## [1] 1log(x=625,base=5) - log(x=125,base=5)## [1] 1\[\log_{b}{\left({x}^{z}\right)} = z\log_{b}{x};\ con\ x\in\textbf{R}^{+},\ z\in\textbf{R}\ y\ b\in\textbf{R}^{+}\setminus\left\{1\right\}\]
\[\log_{2}{\left({64}^{8}\right)} = 8\log_{2}{64}=8\cdot6=48\]
log(x=64**8, base=2)## [1] 482**48==(64**8)## [1] TRUElog(x=281475000000000, base=2)## [1] 488*log(x=64,base=2)## [1] 48\[\log_{3}{\left({243}^{27}\right)} = 27\log_{3}{243}=27\cdot5=135\]
log(x=243**27, base=3)## [1] 1353**135==(243**27)## [1] TRUElog(x=2.578513e+64,base=3)## [1] 13527*log(x=243,base=3)## [1] 135\[\log_{5}{\left({625}^{25}\right)} = 25\log_{5}{625}=25\cdot4=100\]
log(x=625**25, base=5)## [1] 1005**100==(625**25)## [1] TRUElog(x=7.888609e+69, base=5)## [1] 10025*log(x=625,base=5)## [1] 100library(kader)\[\log_{b}{\sqrt[z]{x}} = \log_{b}{\left({x}^{\frac{1}{z}}\right)} = \frac{1}{z}\log_{b}{x};\ con\ x\in\textbf{R}^{+},\ z\in\textbf{R}\setminus\left\{0\right\}\ y\ b\in\textbf{R}^{+}\setminus\left\{1\right\}\]
\[\log_{2}{\left(\sqrt[3]{64}\right)} = \log_{2}{\left({64}^{\frac{1}{3}}\right)} = \frac{1}{3}\log_{2}{64}=\frac{1}{3}\cdot{6}=\frac{6}{3}=2\]
log(x=64**(1/3), base=2)## [1] 2all.equal(2**2,64**(1/3))## [1] TRUEall.equal(2^2,kader:::cuberoot(64))## [1] TRUElog(x=64^(1/3),base=2)## [1] 2(1/3)*log(x=64,base=2)## [1] 2\[\log_{3}{\sqrt[ 5]{243}} = \log_{3}{\left({243}^{\frac{1}{5}}\right)} = \frac{1}{5}\log_{3}{243}=\frac{1}{5}\cdot{5}=\frac{5}{5}=1\]
log(x=243**(1/5), base=3)## [1] 1all.equal(3**1, 243**(1/5))## [1] TRUElog(x=3, base=3)## [1] 1(1/5)*log(x=243, base=3)## [1] 1\[\log_{5}{\sqrt[3]{15625}} = \log_{5}{\left({15.625}^{\frac{1}{3}}\right)} = \frac{1}{3}\log_{5}{15.625}=\frac{1}{3}\cdot6=\frac{6}{3}=2\]
log(x=15625**(1/3), base=5)## [1] 2all.equal(5**2, (15625**(1/3)))## [1] TRUEall.equal(5^2, kader:::cuberoot(15625))## [1] TRUElog(x=25, base=5)## [1] 2(1/3)*log(x=15625, base=5)## [1] 2\[\frac{\log_{c}{\left({x}\right)}}{\log_{c}{\left({b}\right)}} = \log_{b}{\left({x}\right)};\ con\ x\in\textbf{R}^{+},\ z\in\textbf{R}\ y\ b,c\in\textbf{R}^{+}\setminus\left\{1\right\}\]
\[\frac{\log_{2}{\left({64}\right)}}{\log_{2}{\left({8}\right)}} = \log_{8}{\left({64}\right)} = 2\]
log(x=64,base=2)/log(x=8,base=2)## [1] 2all.equal(log(x=64,base=2)/log(x=8,base=2), log(x=64,base=8))## [1] TRUElog(x=64,base=8)## [1] 2\[\frac{\log_{3}{\left({729}\right)}}{\log_{3}{\left({9}\right)}} = \frac{6}{2} = \log_{9}{\left({729}\right)} = 3\]
log(x=729,base=3)/log(x=9,base=3)## [1] 3all.equal(log(x=729,base=3)/log(x=9,base=3), log(x=729,base=9))## [1] TRUElog(x=729, base=9)## [1] 3\[\frac{\log_{5}{\left({625}\right)}}{\log_{5}{\left({25}\right)}} = \frac{4}{2} = \log_{25}{\left({625}\right)} = 2\]
log(x=625,base=5)/log(x=25,base=5)## [1] 2log(x=625,base=25)## [1] 2library(MASS)\[\frac{1}{\log_{x}{\left({b}\right)}} = \log_{b}{\left({x}\right)};\ con\ x\in\textbf{R}^{+},\ z\in\textbf{R}\ y\ b\in\textbf{R}^{+}\setminus\left\{1\right\}\]
\[\frac{1}{\log_{2}{\left({8}\right)}} = \frac{1}{3} = \log_{8}{\left({2}\right)} \leftrightarrow \sqrt[3]{8} = 2\]
fractions(1/log(x=8,base=2))## [1] 1/3fractions(log(x=2,base=8))## [1] 1/3all.equal(1/log(x=8,base=2),log(x=2,base=8))## [1] TRUE\[\frac{1}{\log_{9}{\left({81}\right)}} = \frac{1}{2} = \log_{81}{\left({9}\right)} \leftrightarrow \sqrt[2]{81} = 2\]
fractions(1/log(x=81,base=9))## [1] 1/2fractions(log(x=9,base=81))## [1] 1/2all.equal(1/log(x=81,base=9),log(x=9,base=81))## [1] TRUE\[\frac{1}{\log_{25}{\left({15625}\right)}} = \frac{1}{3} = \log_{15625}{\left({25}\right)} \leftrightarrow \sqrt[3]{15625} = 25\]
fractions(1/log(x=15625,base=25))## [1] 1/3fractions(log(x=25,base=15625))## [1] 1/3all.equal(1/log(x=15625,base=25),log(x=25,base=15625))## [1] TRUE