ĐGNL ĐHQG Hà Nội - Tư duy định lượng - Logarit

A.logab  

B. log_ba

C.ln_ab

D.lnba

A.a < 0, b > 0

B.\[0 < a \ne 1,b < 0\]

C. \[0 < a \ne 1,b > 0\]

D. \[0 < a \ne 1,0 < b \ne 1\]

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.\[{\log _{{a^n}}}b = - n{\log _a}b\]

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.\[{\log _a}{b^n} = n{\log _a}b\]

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.\[{\log _a}b > {\log _a}c\]

B. \[{\log _a}b < {\log _a}c\]

C. \[{\log _a}b < {\log _b}c\]

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. \[\log x < 0 \Leftrightarrow 0 < x < 1\]

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.\[\frac{{10}}{{a - 1}}\]

B. \[\frac{2}{{5(a - 1)}}\]

C.\[\frac{5}{{2a - 2}}\]

D. \[\frac{5}{{2a + 1}}\]

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.\[{\log _3}90 = \frac{{a - 2b + 1}}{{b + 1}}\]

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.\[\frac{{2ab}}{{1 + b}}\]

B. \[\frac{{ab}}{{1 + b}}\]

C. \[\frac{a}{{1 + b}}\]

D. \[\frac{b}{{1 + b}}\]

A.900

B.1350           

C.1050           

D.1200

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. \[{a^3} + {b^3} = 3\left( {8{a^2}b - a{b^2}} \right)\]

A.T=7

B.T=12

C.T=13

D.T=21