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Pilne! Proszę o rozwiazanie. Matematyka liceum



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Odpowiedź :

Trebor5x

a)

[tex]log_{6}(2)+log_{6}(18)=log_{6}(2*18)=2[/tex]

b)

[tex](log_{2}(2\sqrt{2}))^{2}=(log_{2}(8^\frac{1}{2} ))^{2}=(log_{2}(2^\frac{3}{2} ))^{2}=(\frac{3}{2} )^{2}=\frac{9}{4}[/tex]

c)

[tex]log_{5}(1)-log_{5}(25)=log_{5}(\frac{1}{25} )=-2[/tex]

d)

[tex]log_{\sqrt{3} }(16)+log_{\sqrt{3}}(0,125)-log_{\sqrt{3}}(6)=log_{\sqrt{3}}(16*0,125/6)=\\log_{\sqrt{3}}(\frac{1}{3} )=-2[/tex]

5)

[tex]3log_{2}(3)-2log_{2}(5)=log_{2}\frac{3^{3} }{5^{2}} =log_{2}\frac{27}{25}[/tex]

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Basetla

4.

[tex]a) \ log_{6}2 + log_{6}18 = log_{6}(2\cdot18) = log_{6}36 = log_{6}6^{2} = 2\\\\b) \ (log_{a}2\sqrt{2})^{2} = (log_{2}2 + log_{2}\sqrt{2})^{2} = (1+log_{2}2^{\frac{1}{2}})^{2} = (1+\frac{1}{2})^{2} = (\frac{3}{2})^{2} = \frac{9}{4} = 2,25\\\\c) \ log_{5}1 - log_{5}25 = log_{5}\frac{1}{25} =log_{5}5^{-2} = -2[/tex]

[tex]d) \ log_{\sqrt{3}}16 + log_{\sqrt{3}}0,125 - log_{\sqrt{3}}6 = log_{\sqrt{3}}(16\cdot0,125)-log_{\sqrt{3}}6 =\\\\= log_{\sqrt{3}}2 - log_{\sqrt{3}}6 =log_{\sqrt{3}}\frac{2}{6} = log_{\sqrt{3}}\frac{1}{3}} = log_{\sqrt{3}}(\sqrt{3})^{-2} =-2[/tex]

5.

[tex]3log_{2}3 - 2log_{2}5 = log_{2}3^{3}-log_{2}5^{2} = log_{2}27 - log_{2}25 = log_{2}\frac{27}{25} = log_{2}1,08[/tex]

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