Ya hemos estudiado algunas reglas y definiciones importantes con respecto al logaritmo. Ahora aprenderás tres reglas adicionales, que son útiles para resolver ecuaciones logarítmicas. Junto con las mismas restricciones para #\blue{a}# and #\green{b}#, se afirma que #\purple{c}# debería ser positivo.
\[\begin{array}{rcl}\log_{\blue{a}}\left(\green{b}\right)+\log_{\blue{a}}\left(\purple{c}\right)&=&\log_{\blue{a}}\left(\green{b}\cdot \purple{c}\right)\end{array}\]
Ejemplo
\[\begin{array}{lccr}\log_\blue{3}\left(\green{6}\right)+\log_\blue{3}\left(\purple{\sqrt{3}}\right)&=&\log_\blue{3}\left(\green{6}\purple{\sqrt{3}}\right)\end{array}\]
Ten en cuenta que
#\begin{array}{rcl}\blue{a}^{\log_\blue{a}\left(\green{b}\right)+\log_\blue{a}\left(\purple{c}\right)}&=&\blue{a}^{\log_\blue{a}\left(\green{b}\right)}\cdot \blue{a}^{\log_\blue{a}\left(\purple{c}\right)}\\&=&\green{b}\cdot \purple{c}\\&=&\blue{a}^{\log_\blue{a}\left(\green{b}\cdot \purple{c}\right)}\end{array}#
De esto, concluimos que
\[\log_\blue{a}\left(\green{b}\right)+\log_\blue{a}\left(\purple{c}\right) = \log_\blue{a}\left(\green{b}\cdot \purple{c}\right).\]
Ejemplo
#\begin{array}{rcl}\blue{3}^{\log_\blue{3}\left(\green{6}\right)+\log_\blue{3}\left(\purple{\sqrt{3}}\right)}&=&\blue{3}^{\log_\blue{3}\left(\green{6}\right)}\cdot \blue{3}^{\log_\blue{3}\left(\purple{\sqrt{3}}\right)}\\&=&\green{6}\purple{\sqrt{3}}\\&=&\blue{3}^{\log_\blue{3}\left(\green{6}\purple{\sqrt{3}}\right)}\end{array}#
y entonces
\[\log_\blue{3}\left(\green{6}\right)+\log_\blue{3}\left(\purple{\sqrt{3}}\right) = \log_\blue{3}\left(\green{6} \purple{\sqrt{3}}\right).\]
\[\begin{array}{rcl}\log_{\blue{a}}\left(\green{b}\right)-\log_{\blue{a}}\left(\purple{c}\right)&=&\log_{\blue{a}}\left(\frac{\green{b}}{ \purple{c}}\right)\end{array}\]
Ejemplo
\[\begin{array}{lccr}\log_\blue{2}\left(\green{8}\right)-\log_\blue{2}\left(\purple{2}\right)&=&\log_\blue{2}\left(4\right)\end{array}\]
Ten en cuenta que \begin{array}{rcl}\blue{a}^{\log_\blue{a}\left(\green{b}\right)-\log_\blue{a}\left(\purple{c}\right)}&=&\dfrac{\blue{a}^{\log_\blue{a}\left(\green{b}\right)}}{ \blue{a}^{\log_\blue{a}\left(\purple{c}\right)}}\\&=&\dfrac{\green{b}}{\purple{c}}\\&=&\blue{a}^{\log_\blue{a}\left(\frac{\green{b}}{\purple{c}}\right)}\end{array}
De esto concluimos que \[\log_\blue{a}\left(\green{b}\right)-\log_\blue{a}\left(\purple{c}\right)=\log_\blue{a}\left(\frac{\green{b}}{\purple{c}}\right).\]
Ejemplo
#\begin{array}{rcl}\blue{2}^{\log_\blue{2}\left(\green{8}\right)-\log_\blue{2}\left(\purple{2}\right)}&=&\dfrac{\blue{2}^{\log_\blue{2}\left(\green{8}\right)}}{ \blue{2}^{\log_\blue{2}\left(\purple{2}\right)}}\\&=&\dfrac{\green{8}}{\purple{2}}\\&=&\blue{2}^{\log_\blue{2}\left(\frac{\green{8}}{\purple{2}}\right)}\end{array}#
y entonces
\[\log_\blue{2}\left(\green{8}\right)-\log_\blue{2}\left(\purple{2}\right)=\log_\blue{2}\left(\frac{\green{8}}{\purple{2}}\right)=\log_\blue{2}\left(4\right).\]
\[\begin{array}{rcl}\purple{n}\cdot \log_{\blue{a}}\left(\green{b}\right)&=&\log_{\blue{a}}\left(\green{b}^\purple{n}\right)\end{array}\]
Ejemplo
\[\begin{array}{lccr}\purple{3}\cdot \log_\blue{2}\left(\green{4}\right)&=&\log_\blue{2}\left(\green{4}^\purple{3}\right)\\&=&\log_\blue{2}\left(64\right)\end{array}\]
Ten en cuenta que \begin{array}{rcl}\blue{a}^{\purple{n}\cdot \log_\blue{a}\left(\green{b}\right)}&=&\left(\blue{a}^{\log_\blue{a}\left(\green{b}\right)}\right)^\purple{n}\\&=&\green{b}^\purple{n}\\&=&\blue{a}^{\log_\blue{a}\left(\green{b}^\purple{n}\right)}\end{array} De esto, concluimos que \[\purple{n}\cdot \log_\blue{a}\left(\green{b}\right) = \log_\blue{a}\left(\green{b}^\purple{n}\right).\]
Ejemplo
#\begin{array}{rcl}\blue{2}^{\purple{3}\cdot \log_\blue{2}\left(\green{4}\right)}&=&\left(\blue{2}^{\log_\blue{2}\left(\green{4}\right)}\right)^\purple{3}\\&=&\green{4}^\purple{3}\\&=&\blue{2}^{\log_\blue{2}\left(\green{4}^\purple{3}\right)}\end{array}#
y entonces
\[\purple{3}\cdot \log_\blue{2}\left(\green{4}\right) = \log_\blue{2}\left(\green{4}^\purple{3}\right)=\log_\blue{2}\left(64\right).\]
Con la ayuda de estas reglas podemos resolver aún más preguntas con logaritmos.
Simplifica la siguiente expresión a un logaritmo.
#3-2\cdot\log_{4}\left(4\right)#
#\log_4(4)#
\(\begin{array}{rcl}
3-2\cdot\log_{4}\left(4\right)&=&\log_4\left(4^3\right)-2\cdot\log_4\left(4\right)\\
&&\blue{b=\log_a\left(a^b\right)}\\
&=&\log_4\left(64\right)-2\cdot\log_4\left(4\right)\\
&&\blue{\text{se calculó \(4^3\)}}\\
&=&\log_4\left(64\right)-\log_4\left(4^2\right)\\
&&\blue{c\cdot\log_a\left(b\right)=\log_a\left(b^c\right)}\\
&=&\log_4\left(64\right)-\log_4\left(16\right)\\
&&\blue{\text{se calculó \(4^2\)}}\\
&=&\log_4\left(\frac{64}{16}\right)\\
&&\blue{\log_a\left(b\right)-\log_a\left(c\right)=\log_a\left(\frac{b}{c}\right)}\\
\end{array}\)
Reglas para logaritmos
