Oblicz ...- Zadanie 4.1.: Prosto do matury 2. Zakres podstawowy. Po gimnazjum

Rozwiązanie

$lo g_{a} c = b$ wtedy i tylko wtedy, gdy $a^{b} = c$ , ( $c > 0$ , $a > 0$ , $a \neq = 1$ ).

$a)$ $lo g_{5} 25 = 2$ , bo $5^{2} = 25$

$lo g_{7} 343 = 3$ , bo $7^{3} = 343$

$lo g_{6} 1296 = 4$ , bo $6^{4} = 1296$

$lo g_{3} 729 = 6$ , bo $3^{6} = 729$

$lo g_{2} 1024 = 10$ , bo $2^{10} = 1024$

$lo g_{4} 64 = 3$ , bo $4^{3} = 64$

$b)$ $lo g_{8} 2 = \frac{1}{3}$ , bo $8^{\frac{1}{3}} = (2^{3})^{\frac{1}{3}} = 2$

$lo g_{81} 3 = \frac{1}{4}$ , bo $81^{\frac{1}{4}} = (3^{4})^{\frac{1}{4}} = 3$

$lo g_{27} 9 = \frac{2}{3}$ , bo $27^{\frac{2}{3}} = (3^{3})^{\frac{2}{3}} = 3^{2}$

$lo g_{25} (\frac{1}{5}) = - \frac{1}{2}$ , bo $25^{- \frac{1}{2}} = (5^{2})^{- \frac{1}{2}} = 5^{- 1} = \frac{1}{5}$

$lo g_{69} (\frac{1}{69}) = - 1$ , bo $69^{- 1} = \frac{1}{69}$

$lo g_{4} (\frac{1}{64}) = - 3$ , bo $4^{- 3} = (\frac{1}{4})^{3} = \frac{1}{64}$

$c)$ $lo g_{5} 5 = lo g_{5} 5^{\frac{1}{2}} = \frac{1}{2}$

$lo g_{3} (33) = lo g_{3} 9 \cdot 3 = lo g_{3} 27 = lo g_{3} 3^{3} = lo g_{3} 3^{\frac{3}{2}} = \frac{3}{2} = 1 \frac{1}{2}$

$lo g_{2} (4316) = lo g_{2} (4 \cdot 16^{\frac{1}{3}}) = lo g_{2} (4 \cdot (4^{2})^{\frac{1}{3}}) = lo g_{2} (4 \cdot 4^{\frac{2}{3}}) = lo g_{2} 4^{1 + \frac{2}{3}} =$

$= lo g_{2} 4^{1 \frac{2}{3}} = lo g_{2} (2^{2})^{1 \frac{2}{3}} = lo g_{2} (2^{2})^{\frac{5}{3}} = lo g_{2} 2^{\frac{10}{3}} = lo g_{2} 2^{3 \frac{1}{3}} = 3 \frac{1}{3}$

$lo g_{3} (\frac{1}{3} 33) = lo g_{3} (3^{- 1} \cdot 3^{\frac{1}{3}}) = lo g_{3} 3^{- 1 + \frac{1}{3}} = lo g_{3} 3^{- \frac{2}{3}} = - \frac{2}{3}$

$lo g_{2} (0, 125 664) = lo g_{2} (\frac{1}{8} \cdot 64^{\frac{1}{6}}) = lo g_{2} (8^{- 1} \cdot (2^{6})^{\frac{1}{6}}) = lo g_{2} ((2^{3})^{- 1} \cdot 2^{1}) =$

$= lo g_{2} (2^{- 3} \cdot 2^{1}) = lo g_{2} 2^{- 3 + 1} = lo g_{2} 2^{- 2} = - 2$

$lo g 0, 001 = lo g_{10} 0, 001 = lo g_{10} \frac{1}{1000} = lo g_{10} (\frac{1}{1000})^{\frac{1}{2}} = lo g_{10} \frac{1}{10 ^{3}} =$

$= lo g_{10} (\frac{1}{10})^{3} = lo g_{10} 10^{- 3} = lo g_{10} (10^{- 3})^{\frac{1}{2}} = lo g_{10} 10^{- \frac{3}{2}} =$

$= lo g_{10} 10^{- 1 \frac{1}{2}} = - 1 \frac{1}{2}$

$d)$ $lo g_{2} 32 = lo g_{2} 2^{5} = lo g_{2} ((2^{5})^{2})^{\frac{1}{2}} = lo g_{2} (2^{10})^{\frac{1}{2}} = lo g_{2} 2^{10} = lo g_{2} (2)^{10} =$

$= 10$

$lo g_{6} (\frac{1}{36}) = lo g_{6} (\frac{1}{6 ^{2}}) = lo g_{6} (\frac{1}{6})^{2} = lo g_{6} 6^{- 2} = lo g_{6} ((6^{- 2})^{2})^{\frac{1}{2}} =$

$= lo g_{6} (6^{- 4})^{\frac{1}{2}} = lo g_{6} 6^{- 4} = lo g_{6} (6)^{- 4} = - 4$

$lo g_{33} 27 = lo g_{33} 3^{3} = lo g_{33} ((3^{3})^{3})^{\frac{1}{3}} = lo g_{33} (3^{9})^{\frac{1}{3}} = lo g_{33} 3 3^{9} = lo g_{33} (33)^{9} =$

$= 9$

$lo g_{22} 16 = lo g_{2^{1} \cdot 2^{\frac{1}{2}}} 2^{4} = lo g_{2^{1 + \frac{1}{2}}} 2^{4} = lo g_{2^{1 \frac{1}{2}}} 2^{4} = lo g_{2^{\frac{3}{2}}} 2^{4} = lo g_{2^{\frac{3}{2}}} ((2^{4})^{\frac{2}{3}})^{\frac{3}{2}} =$

$= lo g_{2^{\frac{3}{2}}} (2^{\frac{8}{3}})^{\frac{3}{2}} = lo g_{2^{\frac{3}{2}}} (2^{\frac{3}{2}})^{\frac{8}{3}} = \frac{8}{3} = 2 \frac{2}{3}$

$lo g_{(27) 43} 9 = lo g_{3^{3} \cdot 3^{\frac{1}{4}}} 3^{2} = lo g_{3^{3 + \frac{1}{4}}} 3^{2} = lo g_{3^{3 \frac{1}{4}}} 3^{2} = lo g_{3^{\frac{13}{4}}} 3^{2} = lo g_{3^{\frac{13}{4}}} 3^{2} =$

$= lo g_{3^{\frac{13}{4}}} ((3^{2})^{\frac{4}{13}})^{\frac{13}{4}} = lo g_{3^{\frac{13}{4}}} (3^{\frac{8}{13}})^{\frac{13}{4}} = lo g_{3^{\frac{13}{4}}} (3^{\frac{13}{4}})^{\frac{8}{13}} = \frac{8}{13}$

$lo g_{7449} (\frac{1}{7}) = lo g_{7 \cdot 49^{\frac{1}{4}}} 7^{- 1} = lo g_{7 \cdot (7^{2})^{\frac{1}{4}}} 7^{- 1} = lo g_{7^{1} \cdot 7^{\frac{2}{4}}} 7^{- 1} = lo g_{7^{1} \cdot 7^{\frac{1}{2}}} 7^{- 1} =$

$= lo g_{7^{1 + \frac{1}{2}}} 7^{- 1} = lo g_{7^{1 \frac{1}{2}}} 7^{- 1} = lo g_{7^{\frac{3}{2}}} 7^{- 1} = lo g_{7^{\frac{3}{2}}} ((7^{- 1})^{\frac{2}{3}})^{\frac{3}{2}} =$

$= lo g_{7^{\frac{3}{2}}} (7^{- \frac{2}{3}})^{\frac{3}{2}} = lo g_{7^{\frac{3}{2}}} (7^{\frac{3}{2}})^{- \frac{2}{3}} = - \frac{2}{3}$

Agnieszka Sermak

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