Zamień na logarytm przy podstawie...- Zadanie 4.80: Matematyka i przykłady jej zastosowań 1. Zakres podstawowy i rozszerzony

Rozwiązanie

Zmiana podstawy logarytmu

Jeżeli $a > 0, a \neq = 1$ , $b > 0$ i $c > 0, c \neq = 1$ to

$lo g_{a} b = \frac{lo g _{c} b}{lo g _{c} a}$

$lo g_{4} 3 = \frac{lo g _{2} 3}{lo g _{2} 4} = \frac{lo g _{2} 3}{lo g _{2} 2 ^{2}} = \frac{lo g _{2} 3}{2 lo g _{2} 2} = \frac{lo g _{2} 3}{2} = \frac{1}{2} lo g_{2} 3 = lo g_{2} 3^{\frac{1}{2}} = lo g_{2} 3$

$lo g_{8} 5 = \frac{lo g _{2} 5}{lo g _{2} 8} = \frac{lo g _{2} 5}{lo g _{2} 2 ^{3}} = \frac{lo g _{2} 5}{3 lo g _{2} 2} = \frac{lo g _{2} 5}{3} = \frac{1}{3} lo g_{2} 5 = lo g_{2} 5^{\frac{1}{3}} = lo g_{2} 35$

$lo g_{\frac{1}{2}} 3 = \frac{lo g _{2} 3}{lo g _{2} \frac{1}{2}} = \frac{lo g _{2} 3}{lo g _{2} 2 ^{- 1}} = \frac{lo g _{2} 3}{- lo g _{2} 2} = \frac{lo g _{2} 3}{- 1} = - lo g_{2} 3 = lo g_{2} 3^{- 1} = lo g_{2} \frac{1}{3}$

$lo g_{2} 6 = \frac{lo g _{2} 6}{lo g _{2} 2} = \frac{lo g _{2} 6}{lo g _{2} 2 ^{\frac{1}{2}}} = \frac{lo g _{2} 6}{\frac{1}{2} lo g _{2} 2} = \frac{lo g _{2} 6}{\frac{1}{2}} = 2 lo g_{2} 6 = lo g_{2} 6^{2} = lo g_{2} 36$

$lo g_{\frac{1}{2}} 7 = \frac{lo g _{2} 7}{lo g _{2} \frac{1}{2}} = \frac{lo g _{2} 7}{lo g _{2} 2 ^{- \frac{1}{2}}} = \frac{lo g _{2} 7}{- \frac{1}{2} lo g _{2} 2} = \frac{lo g _{2} 7}{- \frac{1}{2}} = - 2 lo g_{2} 7 = lo g_{2} 7^{- 2} = lo g_{2} \frac{1}{49}$

$lo g_{32} 9 = \frac{lo g _{2} 9}{lo g _{2} 32} = \frac{lo g _{2} 9}{lo g _{2} 2 ^{5}} = \frac{lo g _{2} 9}{5} = \frac{1}{5} lo g_{2} 9 = lo g_{2} 9^{\frac{1}{5}} = lo g_{2} 59$

$lo g_{64} 10 = \frac{lo g _{2} 10}{lo g _{2} 64} = \frac{lo g _{2} 10}{lo g _{2} 2 ^{6}} = \frac{lo g _{2} 10}{6} = \frac{1}{6} lo g_{2} 10 = lo g_{2} 1 0^{\frac{1}{6}} = lo g_{2} 610$

$lo g_{16} 10 = \frac{lo g _{2} 10}{lo g _{2} 16} = \frac{lo g _{2} 10}{lo g _{2} 2 ^{4}} = \frac{lo g _{2} 10}{4} = \frac{1}{4} lo g_{2} 10 = lo g_{2} 1 0^{\frac{1}{4}} = lo g_{2} 410$