Utilizăm proprietățile logaritmilor:
[tex]6) \log_{2}12+\log_{2}14-\log_{2}21=\log_{2}(2^2\cdot3)+\log_{2}(2\cdot7)-\log_{2}(3\cdot7) = \log_{2}2^2+\log_{2}3+\log_{2}2+\log_{2}7-(\log_{2}3+\log_{2}7)=2\log_{2}2+\log_{2}3+1+\log_{2}7-\log_{2}3-\log_{2}7 = 2+1 = 3[/tex]
[tex]8) \ \lg\dfrac{2}{1}+\lg\dfrac{3}{2}+...\lg\dfrac{10}{9} = \lg\bigg(\dfrac{\not2}{1} \cdot \dfrac{\not3}{\not2} \cdot \dfrac{\not4}{\not3} \cdot ... \cdot \dfrac{\not9}{\not8} \cdot \dfrac{10}{\not9}\bigg) = \lg\dfrac{10}{1} = \log_{10}10 = 1 \in \Bbb{N}[/tex]
[tex]9) \ 1331^{\dfrac{2}{3}} = (11^3)^{\dfrac{2}{3}} = 11^{3 \cdot\dfrac{2}{3}} = 11^2 = 121 \in \Bbb{N}[/tex]
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Utilizăm proprietățile logaritmilor:
[tex]6) \log_{2}12+\log_{2}14-\log_{2}21=\log_{2}(2^2\cdot3)+\log_{2}(2\cdot7)-\log_{2}(3\cdot7) = \log_{2}2^2+\log_{2}3+\log_{2}2+\log_{2}7-(\log_{2}3+\log_{2}7)=2\log_{2}2+\log_{2}3+1+\log_{2}7-\log_{2}3-\log_{2}7 = 2+1 = 3[/tex]
[tex]8) \ \lg\dfrac{2}{1}+\lg\dfrac{3}{2}+...\lg\dfrac{10}{9} = \lg\bigg(\dfrac{\not2}{1} \cdot \dfrac{\not3}{\not2} \cdot \dfrac{\not4}{\not3} \cdot ... \cdot \dfrac{\not9}{\not8} \cdot \dfrac{10}{\not9}\bigg) = \lg\dfrac{10}{1} = \log_{10}10 = 1 \in \Bbb{N}[/tex]
[tex]9) \ 1331^{\dfrac{2}{3}} = (11^3)^{\dfrac{2}{3}} = 11^{3 \cdot\dfrac{2}{3}} = 11^2 = 121 \in \Bbb{N}[/tex]