Rekenregels voor logaritmen

Exponentiële functies en logaritmen: Logaritmen

Rekenregels voor logaritmen

In de afgelopen paragrafen hebben we al enkele belangrijke regels en definities bestudeerd over de logaritme. We bestuderen nu nog drie van deze regels, die erg handig zijn bij het oplossen van logaritmische vergelijkingen. Naast dezelfde restricties voor $\blue{a}$ en $\green{b}$ , geldt ook dat $\purple{c}$ positief moet zijn.

$\begin{array}{rcl}\log_{\blue{a}}\left(\green{b}\right)+\log_{\blue{a}}\left(\purple{c}\right)&=&\log_{\blue{a}}\left(\green{b}\cdot \purple{c}\right)\end{array}$

Voorbeeld

$\begin{array}{lccr}\log_\blue{3}\left(\green{6}\right)+\log_\blue{3}\left(\purple{\sqrt{3}}\right)&=&\log_\blue{3}\left(\green{6}\purple{\sqrt{3}}\right)\end{array}$

We merken op dat er geldt

$\begin{array}{rcl}\blue{a}^{\log_\blue{a}\left(\green{b}\right)+\log_\blue{a}\left(\purple{c}\right)}&=&\blue{a}^{\log_\blue{a}\left(\green{b}\right)}\cdot \blue{a}^{\log_\blue{a}\left(\purple{c}\right)}\\&=&\green{b}\cdot \purple{c}\\&=&\blue{a}^{\log_\blue{a}\left(\green{b}\cdot \purple{c}\right)}\end{array}$ Hieruit concluderen we dat

$\log_\blue{a}\left(\green{b}\right)+\log_\blue{a}\left(\purple{c}\right) = \log_\blue{a}\left(\green{b}\cdot \purple{c}\right).$

Voorbeeld

$\begin{array}{rcl}\blue{3}^{\log_\blue{3}\left(\green{6}\right)+\log_\blue{3}\left(\purple{\sqrt{3}}\right)}&=&\blue{3}^{\log_\blue{3}\left(\green{6}\right)}\cdot \blue{3}^{\log_\blue{3}\left(\purple{\sqrt{3}}\right)}\\&=&\green{6}\purple{\sqrt{3}}\\&=&\blue{3}^{\log_\blue{3}\left(\green{6}\purple{\sqrt{3}}\right)}\end{array}$ en dus

$\log_\blue{3}\left(\green{6}\right)+\log_\blue{3}\left(\purple{\sqrt{3}}\right) = \log_\blue{3}\left(\green{6} \purple{\sqrt{3}}\right).$

$\begin{array}{rcl}\log_{\blue{a}}\left(\green{b}\right)-\log_{\blue{a}}\left(\purple{c}\right)&=&\log_{\blue{a}}\left(\frac{\green{b}}{ \purple{c}}\right)\end{array}$

Voorbeeld

$\begin{array}{lccr}\log_\blue{2}\left(\green{8}\right)-\log_\blue{2}\left(\purple{2}\right)&=&\log_\blue{2}\left(4\right)\end{array}$

We merken op dat er geldt

$\begin{array}{rcl}\blue{a}^{\log_\blue{a}\left(\green{b}\right)-\log_\blue{a}\left(\purple{c}\right)}&=&\dfrac{\blue{a}^{\log_\blue{a}\left(\green{b}\right)}}{ \blue{a}^{\log_\blue{a}\left(\purple{c}\right)}}\\&=&\dfrac{\green{b}}{\purple{c}}\\&=&\blue{a}^{\log_\blue{a}\left(\frac{\green{b}}{\purple{c}}\right)}\end{array}$

Hieruit concluderen we dat

$\log_\blue{a}\left(\green{b}\right)-\log_\blue{a}\left(\purple{c}\right)=\log_\blue{a}\left(\frac{\green{b}}{\purple{c}}\right).$

Voorbeeld

$\begin{array}{rcl}\blue{2}^{\log_\blue{2}\left(\green{8}\right)-\log_\blue{2}\left(\purple{2}\right)}&=&\dfrac{\blue{2}^{\log_\blue{2}\left(\green{8}\right)}}{ \blue{2}^{\log_\blue{2}\left(\purple{2}\right)}}\\&=&\dfrac{\green{8}}{\purple{2}}\\&=&\blue{2}^{\log_\blue{2}\left(\frac{\green{8}}{\purple{2}}\right)}\end{array}$ en dus

$\log_\blue{2}\left(\green{8}\right)-\log_\blue{2}\left(\purple{2}\right)=\log_\blue{2}\left(\frac{\green{8}}{\purple{2}}\right)=\log_\blue{2}\left(4\right).$

$\begin{array}{rcl}\purple{n}\cdot \log_{\blue{a}}\left(\green{b}\right)&=&\log_{\blue{a}}\left(\green{b}^\purple{n}\right)\end{array}$

Voorbeeld

$\begin{array}{lccr}\purple{3}\cdot \log_\blue{2}\left(\green{4}\right)&=&\log_\blue{2}\left(\green{4}^\purple{3}\right)\\&=&\log_\blue{2}\left(64\right)\end{array}$

We merken op dat er geldt

$\begin{array}{rcl}\blue{a}^{\purple{n}\cdot \log_\blue{a}\left(\green{b}\right)}&=&\left(\blue{a}^{\log_\blue{a}\left(\green{b}\right)}\right)^\purple{n}\\&=&\green{b}^\purple{n}\\&=&\blue{a}^{\log_\blue{a}\left(\green{b}^\purple{n}\right)}\end{array}$ Hieruit concluderen we dat

$\purple{n}\cdot \log_\blue{a}\left(\green{b}\right) = \log_\blue{a}\left(\green{b}^\purple{n}\right).$

Voorbeeld

$\begin{array}{rcl}\blue{2}^{\purple{3}\cdot \log_\blue{2}\left(\green{4}\right)}&=&\left(\blue{2}^{\log_\blue{2}\left(\green{4}\right)}\right)^\purple{3}\\&=&\green{4}^\purple{3}\\&=&\blue{2}^{\log_\blue{2}\left(\green{4}^\purple{3}\right)}\end{array}$ en dus

$\purple{3}\cdot \log_\blue{2}\left(\green{4}\right) = \log_\blue{2}\left(\green{4}^\purple{3}\right)=\log_\blue{2}\left(64\right).$

Met behulp van deze regels kunnen we nog meer vragen met logaritmen oplossen.

Herleid de volgende uitdrukking tot een logaritme.

$4-2\cdot\log_{5}\left(4\right)$

$\log_5({{625}\over{16}})$

$\begin{array}{rcl} 4-2\cdot\log_{5}\left(4\right)&=&\log_5\left(5^4\right)-2\cdot\log_5\left(4\right)\\ &&\blue{b=\log_a\left(a^b\right)}\\ &=&\log_5\left(625\right)-2\cdot\log_5\left(4\right)\\ &&\blue{\text{macht uitwerken}}\\ &=&\log_5\left(625\right)-\log_5\left(4^2\right)\\ &&\blue{c\cdot\log_a\left(b\right)=\log_a\left(b^c\right)}\\ &=&\log_5\left(625\right)-\log_5\left(16\right)\\ &&\blue{\text{macht uitwerken}}\\ &=&\log_5\left(\frac{625}{16}\right)\\ &&\blue{\log_a\left(b\right)-\log_a\left(c\right)=\log_a\left(\frac{b}{c}\right)}\\ \end{array}$

Nieuw voorbeeld