4) a)
[tex]\it log_2 5\ \textgreater \ log_2 4 \Longrightarrow log_2 5\ \textgreater \ 2 \ \ \ \ (*)
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log_3 5\ \textless \ log_3 9 \Longrightarrow log_3 5 \ \textless \ 2 \ \ \ \ (**)
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(*),\ (**) \Longrightarrow log_2 5 \ \textgreater \ log_3 5[/tex]
b)
[tex]\it log_{\dfrac{1}{2}}\dfrac{1}{3} =\dfrac{ log_2 \dfrac{1}{3}}{log_2 \dfrac{1}{2}}=\dfrac{log_2 3^{-1}}{log_2 2^{-1}}=\dfrac{-1\cdot log_2 3}{-1\cdot log_2 2}=\dfrac{log_2 3}{log_2 2} =\dfrac{log_2 3}{1} =log_2 3[/tex]
[tex]\it log_{0,1} \dfrac{1}{3} = log_{\dfrac{1}{10}} \dfrac{1}{3}=\dfrac{lg \dfrac{1}{3}}{lg\dfrac{1}{10}} = \dfrac{lg 3^{-1}}{lg 10^{-1}} =\dfrac{-lg3}{-lg10}=\dfrac{lg3}{1}=lg3[/tex]
[tex]\it log_2 3 \ \textgreater \ log_2 2 \Longrightarrow log_2 3\ \textgreater \ 1\ \ \ \ (*)
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lg3\ \textless \ lg10 \Longrightarrow lg3\ \textless \ 1 \ \ \ \ (**)
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(*),\ (**) \Longrightarrow log_2 3\ \textgreater \ lg3 \Longrightarrow log_{\dfrac{1}{2}}\dfrac{1}{3} \ \textgreater \ log_{0,1}\dfrac{1}{3}[/tex]
7a) lg1 < lg3 < lg10 ⇒ 0 < lg3 < 1
b)
[tex]\it log_{\frac{1}{3}}10 =\dfrac{log_310}{log_3 \dfrac{1}{3}}=\dfrac{log_310}{log_3 3^{-1}} =\dfrac{log_3 10}{-1\cdot log_3 3} = -log_3 10[/tex]
[tex]\it -log_327\ \textless \ -log_310\ \textless \ -log_39 \Longrightarrow -3\ \textless \ -log_310\ \textless \ -2[/tex]
