Przedstaw wyrażenie w postaci...- Zadanie 1: MATeMAtyka 3. Zakres podstawowy. Edycja 2024

Rozwiązanie

Przypomnijmy, że:

Jeśli a, b, x są liczbami dodatnimi oraz a≠1, b≠1, to

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

$lo g_{16} 3 = \frac{lo g _{2} 3}{lo g _{2} 16} = \frac{lo g _{2} 3}{lo g _{2} 2 ^{4}} = \frac{lo g _{2} 3}{4} = \frac{1}{4} \cdot lo g_{2} 3 = lo g_{2} 3^{\frac{1}{4}} = lo g_{2} 43$

$lo g_{0, 5} 7 = \frac{lo g _{2} 7}{lo g _{2} 0 , 5} = \frac{lo g _{2} 7}{lo g _{2} \frac{1}{2}} = \frac{lo g _{2} 7}{lo g _{2} 2 ^{- 1}} = - lo g_{2} 7 = lo g_{2} 7^{- 1} = lo g_{2} \frac{1}{7}$

$lo g_{2} 11 = \frac{lo g _{2} 11}{lo g _{2} 2} = \frac{lo g _{2} 11}{lo g _{2} 2 ^{\frac{1}{2}}} = 2 \cdot lo g_{2} 11 = lo g_{2} 11^{2} = lo g_{2} 121$

$lo g_{\frac{1}{8}} 9 = \frac{lo g _{2} 9}{lo g _{2} \frac{1}{8}} = \frac{lo g _{2} 9}{lo g _{2} ( \frac{1}{2} ) ^{3}} = \frac{lo g _{2} 9}{lo g _{2} 2 ^{- 3}} = - \frac{1}{3} \cdot lo g_{2} 9 = lo g_{2} 9^{- \frac{1}{3}}$

$lo g_{4} 6 + lo g_{8} 6 = \frac{lo g _{2} 6}{lo g _{2} 4} + \frac{lo g _{2} 6}{lo g _{2} 8} = \frac{lo g _{2} 6}{lo g _{2} 2 ^{2}} + \frac{lo g _{2} 6}{lo g _{2} 2 ^{3}} = \frac{lo g _{2} 6}{2} + \frac{lo g _{2} 6}{3} =$

$= \frac{1}{2} \cdot lo g_{2} 6 + \frac{1}{3} \cdot lo g_{2} 6 = lo g_{2} 6^{\frac{1}{2}} + lo g_{2} 6^{\frac{1}{3}} = lo g_{2} (6^{\frac{1}{2}} \cdot 6^{\frac{1}{3}}) = lo g_{2} 6^{\frac{5}{6}}$

$lo g_{\frac{1}{2}} 3 + lo g_{\frac{1}{4}} 3 + lo g_{0, 125} 3 = \frac{lo g _{2} 3}{lo g _{2} \frac{1}{2}} + \frac{lo g _{2} 3}{lo g _{2} \frac{1}{4}} + \frac{lo g _{2} 3}{lo g _{2} 0 , 125} =$

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