Reglas para logaritmos

Funciones exponenciales y logaritmos: Funciones logarítmicas

Reglas para logaritmos

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.

$2-4\cdot\log_{3}\left(2\right)$

$\log_3({{9}\over{16}})$

$\begin{array}{rcl} 2-4\cdot\log_{3}\left(2\right)&=&\log_3\left(3^2\right)-4\cdot\log_3\left(2\right)\\ &&\blue{b=\log_a\left(a^b\right)}\\ &=&\log_3\left(9\right)-4\cdot\log_3\left(2\right)\\ &&\blue{\text{se calculó $3^2$}}\\ &=&\log_3\left(9\right)-\log_3\left(2^4\right)\\ &&\blue{c\cdot\log_a\left(b\right)=\log_a\left(b^c\right)}\\ &=&\log_3\left(9\right)-\log_3\left(16\right)\\ &&\blue{\text{se calculó $2^4$}}\\ &=&\log_3\left(\frac{9}{16}\right)\\ &&\blue{\log_a\left(b\right)-\log_a\left(c\right)=\log_a\left(\frac{b}{c}\right)}\\ \end{array}$

Nuevo ejemplo