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Теория: 00 Теория

Задание

Свойства логарифмов

Правило

 

Основное свойство логарифма

\(\displaystyle \color{green}{a}^{\log_\color{green}{a} \color{blue}{b}} = \color{blue}{b}\)

(\(\displaystyle a>0{\small,}\) \(\displaystyle b>0{\small,}\)\(\displaystyle a\cancel=1)\)

 

Вынесение показателя степени из логарифма

  • \(\displaystyle \log_a b^\color{red}n =\color{red}n\log_a b\)

(\(\displaystyle a>0{\small,}\)\(\displaystyle b>0{\small,}\)\(\displaystyle a\cancel=1)\)

  • \(\displaystyle {\log_{a^\color{red}n} b =\frac{1}{\color{red}n}\log_a b}\)

(\(\displaystyle a>0{\small,}\)\(\displaystyle b>0{\small,}\)\(\displaystyle a\cancel=1{\small,}\)\(\displaystyle n\cancel=0)\)

Сумма и разность логарифмов с одинаковыми основаниями

  • \(\displaystyle \log_\color{red}{a} \color{blue}x + \log_\color{red}{a} \color{green}{y}=\log_\color{red}{a} (\color{blue}x\cdot \color{green}{y}){\small,}\)
  • \(\displaystyle \log_\color{red}{a} \color{blue}x - \log_\color{red}{a} \color{green}{y}=\log_\color{red}{a} \left(\frac{\color{blue}x}{ \color{green}{y}}\right)\)

\(\displaystyle (a>0{\small,}\)\(\displaystyle a\cancel=1{\small,}\)\(\displaystyle x>0{\small,}\)\(\displaystyle y>0)\)

 

Переход к новому основанию

  • \(\displaystyle \frac{1} { \log_{\color{green}{a}}\color{blue}{b}}=\log_{\color{blue}{b}}\color{green}{a}\)

(\(\displaystyle a>0{\small,}\) \(\displaystyle b>0{\small,}\)\(\displaystyle a\cancel=1{\small,}\)\(\displaystyle b\cancel=1)\)

  • \(\displaystyle \frac{\log_{\color{red}{c}}\color{green}{a}} { \log_{\color{red}{c}}\color{blue}{b}}=\log_{\color{blue}{b}}\color{green}{a}\)

\(\displaystyle (a>0{\small,}\)\(\displaystyle b>0{\small,}\)\(\displaystyle c>0{\small,}\)\(\displaystyle b\cancel=1{\small,}\)\(\displaystyle c\cancel=1)\)

Решение