Twierdzenie o zmianie podstawy logarytmu Dla $a,b,c>0$  i  $a\ne 1$, $b\ne 1$ zachodzą następująca równość: ${\log }_{b}c=\frac{{\log }_{a}c}{{\log }_{a}b}$

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a)

Mamy dane:

${\log }_{0,1}7$

$c=10$

${\log }_{0,1}7=\frac{{\log }_{10}7}{{\log }_{10}0,1}=\frac{{\log }_{10}7}{-1}=-{\log }_{10}7=$      

$={\log }_{10}{7}^{-1}={\log }_{10}\frac{1}{7}$

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b)

Mamy dane:

${\log }_{8}3$

$c=2$

${\log }_{8}3=\frac{{\log }_{2}3}{{\log }_{2}8}=\frac{{\log }_{2}3}{3}=\frac{1}{3}{\log }_{2}3=$      

$={\log }_2{3}^{\frac{1}{3}}={\log }_2\sqrt[2]{3}$

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c)

Mamy dane:

${\log }_{7}11$

$c=49$

${\log }_{7}11=\frac{{\log }_{49}11}{{\log }_{49}7}=\frac{{\log }_{49}11}{\frac{1}{2}}=2{\log }_{49}11={\log }_{49}121$

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d)

Mamy dane:

${\log }_{}625$

$c=0,1$

${\log }_{}625={\log }_{10}625=\frac{{\log }_{0,1}625}{{\log }_{0,1}10}=\frac{{\log }_{0,1}625}{-1}=$

$=-{\log }_{0,1}625={\log }_{0,1}\frac{1}{625}$

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e)

Mamy dane:

${\log }_{2}6$

$c=\frac{1}{2}$

${\log }_{2}6=\frac{{\log }_{\frac{1}{2}}6}{{\log }_{\frac{1}{2}}2}=\frac{{\log }_{\frac{1}{2}}6}{-1}=-{\log }_{\frac{1}{2}}6={\log }_{\frac{1}{2}}\frac{1}{6}$

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f)

Mamy dane:

${\log }_{\frac{1}{3}}12$

$c=3$

${\log }_{\frac{1}{3}}12=\frac{{\log }_{3}12}{{\log }_3\frac{1}{3}}=\frac{{\log }_{3}12}{-1}=-{\log }_{3}12={\log }_3\frac{1}{12}$