Odpowiedź :
[tex]log_{5}25 = log_{5}5^{2} = 2\\\\log_{3}27 = log_{3}3^{3} = 3\\\\log_{6}\frac{1}{6} = log_{6} 6^{-1} = -1\\\\log_{7}\frac{1}{49} = log_{7}7^{-2} = -2\\\\log_{9} 3 = log_{9}9^{\frac{1}{2}} = \frac{1}{2}\\\\log_{1000} 10 = log_{1000} 1000^{\frac{1}{3}} = \frac{1}{3}\\\\log_{5} \sqrt{5} = log_{5} 5^{\frac{1}{2}} = \frac{1}{2}\\\\log_{7}\sqrt[4]{7} = log_{7} 7^{\frac{1}{4}} = \frac{1}{4}\\\\log_{3}\sqrt[3]{3} = log_{3} 3^{\frac{1}{3}} = \frac{1}{3}[/tex]
[tex]log_{6}\sqrt[3]{36} = log_{6} (6^{2})^{\frac{1}{3}} = log_{6} 6^{\frac{2}{3}} = \frac{2}{3}\\\\log_{2}16 = log_{2}2^{4} = 4\\\\log_{8}2 = log_{8}8^\frac{1}{3}} = \frac{1}{3}[/tex]
Odpowiedź:
Korzystam ze wzorów:
㏒ₐ a = 1
㏒ₐ bˣ = x㏒ₐ b
㏒ₐ b = x z def. log. ⇒ aˣ = b
Szczegółowe wyjaśnienie:
