Explicație pas cu pas:
a)
[tex]10 = log_{2}({2}^{10}) = log_{2}(1048) < log_{2}(2012) < log_{2}(2048) = log_{2}({2}^{11}) = 11 \\ [/tex]
[tex]\implies \bf \Big[log_{2}(2012)\Big] = 10[/tex]
b)
[tex]3 = lg({10}^{3}) = lg(1000) < lg(2012) < lg(10000) = lg({10}^{4}) = 4 \\ [/tex]
[tex]\implies \bf \Big[lg(2012)\Big] = 3[/tex]
c)
[tex]4 = log_{3}({3}^{4}) = log_{3}(81) < log_{3}(100) < log_{3}(243) = log_{3}({3}^{5}) = 5 \\ [/tex]
[tex]\implies \bf \Big[log_{3}(100)\Big] = 4[/tex]
d)
[tex]log_{ \frac{1}{2} }(9) = - log_{2}(9)[/tex]
[tex]3 = log_{2}( {2}^{3} ) = log_{2}(8) < log_{2}(9) < log_{2}(16) = log_{2}({2}^{4}) = 4 \\ - 4 < - log_{2}(9) < - 3 \implies - 4 < log_{ \frac{1}{2} }(9) < - 3[/tex]
[tex]\implies \bf \Big[log_{ \frac{1}{2} }(9)\Big] = - 4 \\ [/tex]
e)
[tex]log_{0.1}(11) = log_{ \frac{1}{10} }(11) = - log_{10}(11) \\ [/tex]
[tex]1 = log_{10}(10) < log_{10}(11) < log_{10}(100) = log_{10}( {10}^{2} ) = 2 \\ - 2 < - log_{10}(11) < - 1 \implies - 2 < - log_{0.1}(11) < - 1[/tex]
[tex]\implies \bf \Big[log_{0.1}(9)\Big] = - 2 \\ [/tex]
f)
[tex]5 = log_{3}({3}^{5}) = log_{3}(243) < log_{3}(321) < log_{3}(729) = log_{3}({3}^{6}) = 6 \\ [/tex]
[tex]\implies \bf \Big[log_{3}(321)\Big] = 5[/tex]
