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. \[{\log _{{a^2}}}\left( {ab} \right) = \frac{1}{2}{\log _a}b\]