Oblicz ...- Zadanie 36.: Prosto do matury 2. Zakres podstawowy

Rozwiązanie

Wzory, które będą nam potrzebne do rozwiązania tego zadania:

$a^{l o g_{a} b} = b$
$k \cdot lo g_{a} b = lo g_{a} b^{k}$

$a)$ $3^{l o g_{3} 18} = 18$

$b)$ $25^{l o g_{5} 7} = (5^{2})^{l o g_{5} 7} = 5^{2 l o g_{5} 7} = 5^{l o g_{5} 7^{2}} = 5^{l o g_{5} 49} = 49$

$c)$ $8^{1 - l o g_{8} 4} = 8^{1} \cdot 8^{- l o g_{8} 4} = 8 \cdot 8^{l o g_{8} 4^{- 1}} = 8 \cdot 8^{l o g_{8} (\frac{1}{4})} = 8 \cdot \frac{1}{4} = \frac{8}{4} = 2$

$d)$ $2^{2 - l o g_{4} 9} = 2^{2} \cdot 2^{- l o g_{4} 9} = 4 \cdot (2^{2 \cdot \frac{1}{2}})^{- l o g_{4} 9} = 4 \cdot ((2^{2})^{\frac{1}{2}})^{- l o g_{4} 9} = 4 \cdot (2^{2})^{- \frac{1}{2} l o g_{4} 9} =$

$= 4 \cdot 4^{l o g_{4} 9^{- \frac{1}{2}}} = 4 \cdot 9^{- \frac{1}{2}} = 4 \cdot (\frac{1}{9})^{\frac{1}{2}} = 4 \cdot \frac{1}{9} = 4 \cdot \frac{1}{9} = 4 \cdot \frac{1}{3} = \frac{4}{3} = 1 \frac{1}{3}$

$e)$ $81^{1 - l o g_{3} 27^{\frac{1}{4}}} = (3^{4})^{1 - l o g_{3} 27^{\frac{1}{4}}} = 3^{4 \cdot (1 - l o g_{3} 27^{\frac{1}{4}})} = 3^{4 - 4 l o g_{3} 3^{\frac{3}{4}}} = 3^{4 - l o g_{3} (27^{\frac{1}{4}})^{4}} =$

$= 3^{4 - l o g_{3} 27^{\frac{1}{4} \cdot 4}} = 3^{4 - l o g_{3} 27} = 3^{4} \cdot 3^{- l o g_{3} 27} = 81 \cdot 3^{l o g_{3} 27^{- 1}} = 81 \cdot 3^{l o g_{3} (\frac{1}{27})} =$

$= 81 \cdot \frac{1}{27} = \frac{81}{27} = 3$

$f)$ $216^{\frac{5 - l o g _{6} 36}{9}} = 216^{\frac{5 - l o g _{6} 6 ^{2}}{9}} = 216^{\frac{5 - 2}{9}} = 216^{\frac{3}{9}} = 216^{\frac{1}{3}} = (6^{3})^{\frac{1}{3}} = 6^{3 \cdot \frac{1}{3}} = 6^{1} = 6$