Bài 27: Hãy tìm lôgarit của mỗi số sau theo cơ số 3:
3; 81; 1; \({1 \over 9};\root 3 \of 3 ;{1 \over {3\sqrt 3 }}\).
Áp dụng \({\log _a}{a^b} = b\,\,\) với \(a > 0;a \ne 1\)
\({\log _3}3 = 1;{\log _3}81 = {\log _3}{3^4} = 4;{\log _3}1 = 0;\)
\({\log _3}{1 \over 9} = {\log _3}{3^{ – 2}} = – 2;\)
\({\log _3}\root 3 \of 3 = {\log _3}{3^{{1 \over 3}}} = {1 \over 3};{\log _3}{1 \over {3\sqrt 3 }} = {\log _3}{3^{{{ – 3} \over 2}}} = – {3 \over 2}\)
Bài 28: Tính \({\log _{{1 \over 5}}}125;{\log _{0,5}}{1 \over 2};{\log _{{1 \over 4}}}{1 \over {64}};{\log _{{1 \over 6}}}36.\)
\({\log _{{1 \over 5}}}125 = {\log _{{1 \over 5}}}{\left( {{1 \over 5}} \right)^{ – 3}} = – 3;\)
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\({\log _{0,5}}{1 \over 2} = {\log _{0,5}}0,5 = 1;\)
\({\log _{{1 \over 4}}}{1 \over {64}} = {\log _{{1 \over 4}}}{\left( {{1 \over 4}} \right)^3} = 3;\)
\({\log _{{1 \over 6}}}36 = {\log _{{1 \over 6}}}{\left( {{1 \over 6}} \right)^{ – 2}} = – 2.\)
Bài 29: Tính \({3^{{{\log }_3}18}};{3^{5{{\log }_3}2}};{\left( {{1 \over 8}} \right)^{{{\log }_2}5}};{\left( {{1 \over {32}}} \right)^{{{\log }_{0,5}}2}}\)
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Áp dụng \({a^{{{\log }_a}b}} = b\left( {a > 0,a \ne 1} \right)\)
\({3^{{{\log }_3}18}} = 18;\) \({3^{5{{\log }_3}2}} = {3^{lo{g_3}{2^5}}} = {2^5} = 32;\)
\({\left( {{1 \over 8}} \right)^{{{\log }_2}5}} = {\left( {{2^{ – 3}}} \right)^{{{\log }_2}5}} = {2^{\left( { – 3} \right){{\log }_2}5}} = {2^{{{\log }_2}{5^{ – 3}}}} \)
\(= {5^{ – 3}} = {1 \over {125}};\)
\({\left( {{1 \over {32}}} \right)^{{{\log }_{0,5}}2}} = {\left( {{{\left( {{1 \over 2}} \right)}^5}} \right)^{{{\log }_{{1 \over 2}}}2}} = {\left( {{1 \over 2}} \right)^{lo{g_{{1 \over 2}}}{2^5}}} = {2^5} = 32;\)
Bài 30: Tìm x, biết:
a) \({\log _5}x = 4;\) b) \({\log _2}\left( {5 – x} \right) = 3;\)
c) \({\log _3}\left( {x + 2} \right) = 3;\) d) \({\log _{{1 \over {16}}}}\left( {0,5 + x} \right) = – 1;\)
a) \({\log _5}x = 4 \Leftrightarrow x = {5^4} = 625.\)
b) \({\log _2}\left( {5 – x} \right) = 3 \Leftrightarrow 5 – x = {2^3} \Leftrightarrow x = – 3\);
c) \({\log _3}\left( {x + 2} \right) = 3 \Leftrightarrow x + 2 = {3^3} \Leftrightarrow x = 25\);
d) \({\log _{{1 \over {6}}}}\left( {0,5 + x} \right) = – 1 \Leftrightarrow 0,5 + x = {\left( {{1 \over {6}}} \right)^{ – 1}}\)
\(\Leftrightarrow x = 6 – 0,5 = 5,5\).
