A.logab
B. logba
C.lnab
D.lnba
A.\[{a^b} = N\]
B.\[{\log _a}N = b\]
C.\[{a^N} = b\]
D.\[{b^N} = a\]
A.\[{\log _a}\left( {bc} \right) = {\log _a}b + {\log _b}c\]
B. \[{\log _a}\frac{b}{c} = {\log _a}b + {\log _a}c\]
C. \[{\log _a}\frac{b}{c} = \frac{{{{\log }_a}b}}{{{{\log }_a}c}}\]
D. \[{\log _a}\left( {bc} \right) = {\log _a}b + {\log _a}c\]
A.\[{\log _a}{b^n} = n{\log _a}b\]
B. \[{\log _a}\sqrt[n]{b} = \frac{1}{n}{\log _a}b\]
C. \[{\log _a}\frac{1}{b} = - {\log _a}b\]
D. \[{\log _a}\sqrt[n]{b} = - n{\log _a}b\]
A.\[{\log _{{a^n}}}b = - n{\log _a}b\]
B. \[{\log _{{a^n}}}b = \frac{1}{n}{\log _a}b\]
C. \[{\log _{{a^n}}}b = - \frac{1}{n}{\log _a}b\]
D. \[{\log _{{a^n}}}b = n{\log _a}b\] Trả lời:
A.\[{\log _a}b > {\log _a}c\]
B. \[{\log _a}b >
C. \[{\log _a}b >
D.\[{\log _a}b > {\log _c}b\]
A.\[{\log _a}1 = 1\]
B. \[{\log _a}a = a\]
C. \[{\log _a}1 = a\]
D. \[{\log _a}a = 1\]
A.\[{\log _a}{a^b} = b\]
B. \[{\log _a}{a^b} = {a^b}\]
C. \[{a^{{{\log }_a}b}} = b\]
D. \[{a^{{{\log }_a}b}} = {\log _a}{a^b}\]
A.\[{2^{{{\log }_2}3}} = {5^{{{\log }_3}5}}\]
B. \[{2^{{{\log }_2}3}} = {5^{{{\log }_5}3}}\]
C. \[{5^{{{\log }_5}3}} = {\log _2}3\]
D. \[{2^{{{\log }_2}4}} = 2\]
A.\[{\log _5}6 = {\log _2}6.{\log _3}6\]
B. \[{\log _5}6 = {\log _5}2 + {\log _5}3\]
C. \[{\log _5}6 = {\log _5}5 + {\log _5}1\]
D. \[{\log _5}6 = {\log _5}2.{\log _5}3\]
A.\[{\log _a}b.{\log _b}c = {\log _a}c\]
B. \[{\log _b}c = \frac{{{{\log }_a}b}}{{{{\log }_a}c}}\]
C. \[{\log _a}b = {\log _c}b - {\log _c}a\]
D. \[{\log _a}b + {\log _b}c = {\log _a}c\]
A.\[{\log _2}3 = - {\log _3}2\]
B. \[{\log _3}2.{\log _3}\frac{1}{2} = 1\]
C. \[{\log _2}3 + {\log _3}2 = 1\]
D. \[{\log _2}3 = \frac{1}{{{{\log }_3}2}}\]
A.\[{\log _{{a^n}}}b = {\log _{{b^n}}}a\]
B. \[{\log _{{a^n}}}b = \frac{1}{{{{\log }_{{b^n}}}a}}\]
C. \[{\log _{{a^n}}}b = {\log _a}\sqrt[n]{b}\]
D. \[{\log _{{a^n}}}b = n{\log _{{b^n}}}a\]
A.2
B.−8
C.−2
D.\(\frac{1}{2}\)
A.\[\frac{3}{4}\]
B. \(\frac{1}{2}\)
C. \[\frac{1}{3}\]
D. \[\frac{5}{6}\]
A.0 < a < 1
B.0 < a< 3
C.a > 3
D.a > 1
A.\[2\log a + \log b\]
B. \[\log a + 2\log b\]
C. \[2\left( {\log a + \log b} \right)\]
D. \[\log a + \frac{1}{2}\log b\]
A.\[P = {\log _2}{\left( {\frac{a}{b}} \right)^2}\]
B. \[P = {\log _2}\left( {\frac{{2a}}{{{b^2}}}} \right)\]
C. \[P = {\log _2}\left( {2a{b^2}} \right)\]
D. \[P = {\log _2}{\left( {ab} \right)^2}\]
A.\[{\log _{{a^2}}}\left( {ab} \right) = \frac{1}{2} + \frac{1}{2}{\log _a}b\]
B. \[{\log _{{a^2}}}\left( {ab} \right) = 2 + {\log _a}b\]
C. \[{\log _{{a^2}}}\left( {ab} \right) = \frac{1}{4}{\log _a}b\]
D. \[{\log _{{a^2}}}\left( {ab} \right) = \frac{1}{2}{\log _a}b\]
A.\[P = \frac{{\sqrt 3 }}{3}\]
B.\[P = \frac{1}{3}\]
C.P=27
D. \[P = 3\sqrt 3 \]
A.\[{\log _{0,5}}a > {\log _{0,5}}b \Leftrightarrow a > b > 0\]
B. \[\log x >
C. \[{\log _2}x > 0 \Leftrightarrow x > 1\]
D. \[{\log _{\frac{1}{3}}}a = {\log _{\frac{1}{3}}}b \Leftrightarrow a = b > 0\]
A.a>1,0<b<1
B.0<a<1,0<b<1
C.0<a<1,b>1
D.a>1,b>1
A.\[{\log _a}b < 1 < {\log _b}a\]
B. \[1 < {\log _a}b < {\log _b}a\]
C. \[{\log _b}a < {\log _a}b < 1\]
D. \[{\log _b}a < 1 < {\log _a}b\]
A.\[a > b > c\;\;\;\]
B.\[c > a > b\]
C.\[c > b > a\]
D.\[b > a > c\]
A.\[P = \frac{{2a + b + 3}}{a}\]
B. \[P = \frac{{a + b + 4}}{a}\]
C. \[P = \frac{{a + b + 3}}{a}\]
D. \[P = \frac{{a + 2b + 3}}{a}\]
A.\[{\log _6}45 = \frac{{2{a^2} - 2ab}}{{ab}}\]
B. \[{\log _6}45 = \frac{{2{a^2} - 2ab}}{{ab + b}}\]
C. \[{\log _6}45 = \frac{{a + 2ab}}{{ab + b}}\]
D. \[{\log _6}45 = \frac{{a + 2ab}}{{ab}}\]
A.\[\frac{{1 - a}}{{a - 2}}\]
B. \[\frac{{2a - 1}}{{a - 2}}\]
C. \[\frac{{a - 1}}{{2a - 2}}\]
D. \[\frac{{1 - 2a}}{{a - 2}}\]
A.\[P = \frac{{ab + 2a + 2}}{b}\]
B. \[P = \frac{{ab - a + 2}}{b}\]
C.\[P = \frac{{ab + a - 2}}{b}\]
D. \[P = \frac{{ab - a - 2}}{b}\]
A.\[\frac{{10}}{{a - 1}}\]
B. \[\frac{2}{{5(a - 1)}}\]
C.\[\frac{5}{{2a - 2}}\]
D. \[\frac{5}{{2a + 1}}\]
A.\[{\log _3}90 = \frac{{a - 2b + 1}}{{b + 1}}\]
B. \[{\log _3}90 = \frac{{a + 2b - 1}}{{b - 1}}\]
C. \[{\log _3}90 = \frac{{2a - b + 1}}{{a + 1}}\]
D. \[{\log _3}90 = \frac{{2a + b - 1}}{{a - 1}}\]
A.\[{a^2}{p^4}\]
B. \[4p + 2\]
C. \[4p + 2a\]
D. \[{p^4} + 2a\]
A.\[{\log _{12}}80 = \frac{{2{a^2} - 2ab}}{{ab + b}}\]
B.\[{\log _{12}}80 = \frac{{a + 2ab}}{{ab}}\]
C. \[{\log _{12}}80 = \frac{{a + 2ab}}{{ab + b}}\]
D. \[{\log _{12}}80 = \frac{{2{a^2} - 2ab}}{{ab}}\]
A.\[{\log _2}7 = \frac{a}{{1 - b}}\]
B.\[{\log _2}7 = \frac{b}{{1 - a}}\]
C. \[{\log _2}7 = \frac{a}{{1 + b}}\]
D. \[{\log _2}7 = \frac{b}{{1 + a}}\]
A.\[2\log \left( {a + 2b} \right) = 5\left( {\log a + \log b} \right)\]
B.\[\log \left( {a + 1} \right) + \log b = 1\]
C. \[\log \frac{{a + 2b}}{3} = \frac{{\log a + \log b}}{2}\]
D. \[5\log \left( {a + 2b} \right) = \log a - \log b\]
A.T=−3
B.T=3
C.T=−1
D.T=1
A.\[P = 2{\ln ^2}a + 1\]
B.\[P = 2{\ln ^2}a + 2\]
C. \[P = 2{\ln ^2}a\]
D. \[P = {\ln ^2}a + 2\]
A.0
B.1
C.\[\ln (\frac{a}{b} + \frac{b}{c} + \frac{c}{d} + \frac{d}{a})\]
D. \[\ln (abcd)\]
A.900
B.1350
C.1050
D.1200
A.\[\frac{{2ab}}{{1 + b}}\]
B. \[\frac{{ab}}{{1 + b}}\]
C. \[\frac{a}{{1 + b}}\]
D. \[\frac{b}{{1 + b}}\]
A.T=7
B.T=12
C.T=13
D.T=21
A.\[\ln \left( {a + 2b} \right) - 2\ln 2 = \ln a + \ln b\]
B. \[\ln \left( {a + 2b} \right) = \frac{1}{2}(\ln a + \ln b)\]
C. \[\ln \left( {a + 2b} \right) - 2\ln 2 = \frac{1}{2}(\ln a + \ln b)\]
D. \[\ln \left( {a + 2b} \right) + 2\ln 2 = \frac{1}{2}(\ln a + \ln b)\]
A.\[{a^3} + {b^3} = 8{a^2}b - a{b^2}\]
B. \[{a^3} + {b^3} = 3\left( {8{a^2}b + a{b^2}} \right)\]
C. \[{a^3} + {b^3} = 3\left( {{a^2}b - a{b^2}} \right)\]
D. \[{a^3} + {b^3} = 3\left( {8{a^2}b - a{b^2}} \right)\]
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