Logarit

$P = {\log _3}240 = \dfrac{{{{\log }_2}240}}{{{{\log }_2}3}} = \dfrac{{{{\log }_2}\left( {{2^4}.3.5} \right)}}{{{{\log }_2}3}} = \dfrac{{{{\log }_2}{2^4} + {{\log }_2}3 + {{\log }_2}5}}{{{{\log }_2}3}} = \dfrac{{a + b + 4}}{a}$

Ta có

$\begin{array}{l}T = 2\ln \sqrt {ex}  - \ln \dfrac{{{e^2}}}{{\sqrt x }} + \ln 3.{\log _3}e{x^2}\\ = 2\ln \left( {{e^{\dfrac{1}{2}}}.{x^{\dfrac{1}{2}}}} \right) - \left( {\ln {e^2} - \ln {x^{\dfrac{1}{2}}}} \right) + \ln 3.\dfrac{{\ln \left( {e.{x^2}} \right)}}{{\ln 3}}\\ = 2\left( {\dfrac{1}{2} + \dfrac{1}{2}\ln x} \right) - \left( {2 - \dfrac{1}{2}\ln x} \right) + \ln e + 2\ln x\\ = 2\left( {\dfrac{1}{2} + \dfrac{1}{2}.2} \right) - \left( {2 - \dfrac{1}{2}.2} \right) + 1 + 2.2 = 7\end{array}$

Có $a = {\log _2}3 \Rightarrow {\log _3}2 = \dfrac{1}{a};b = {\log _5}3 \Rightarrow {\log _3}5 = \dfrac{1}{b}$

${\log _6}45 = \dfrac{{{{\log }_3}45}}{{{{\log }_3}6}} = \dfrac{{{{\log }_3}\left( {{3^2}.5} \right)}}{{{{\log }_3}\left( {2.3} \right)}} = \dfrac{{2 + {{\log }_3}5}}{{{{\log }_3}2 + 1}} = \dfrac{{2 + \dfrac{1}{b}}}{{\dfrac{1}{a} + 1}} = \dfrac{{2ab + a}}{{ab + b}}$

Ta có: \({\log _{10e}}x = \dfrac{1}{{{{\log }_x}10e}} = \dfrac{1}{{{{\log }_x}e + {{\log }_x}10}} = \dfrac{1}{{\dfrac{{\ln e}}{{\ln x}} + \dfrac{{\ln 10}}{{\ln x}}}} = \dfrac{{\ln x}}{{1 + \ln 10}} = \dfrac{{\ln 10.\log x}}{{1 + \ln 10}}\)

Suy ra \({\log _{10e}}x = \dfrac{{ab}}{{1 + b}}\).

Từ công thức ${\log _a}b.{\log _b}c = {\log _a}c \Leftrightarrow {\log _b}c = \dfrac{{{{\log }_a}c}}{{{{\log }_a}b}}\left( {0 < a,b \ne 1;c > 0} \right)$ ta thấy chỉ có đáp án A đúng.

Ta có:

\(\begin{array}{l}{\log _{15}}20 = {\log _{15}}\left( {{2^2}.5} \right)\\ = 2{\log _{15}}2 + {\log _{15}}5\\ = \dfrac{2}{{{{\log }_2}15}} + \dfrac{1}{{{{\log }_5}15}}\\ = \dfrac{2}{{{{\log }_2}3 + {{\log }_2}5}} + \dfrac{1}{{{{\log }_5}3 + {{\log }_5}5}}\\ = \dfrac{2}{{\dfrac{1}{{{{\log }_3}2}} + \dfrac{{{{\log }_3}5}}{{{{\log }_3}2}}}} + \dfrac{1}{{{{\log }_5}3 + 1}}\\ = \dfrac{{2{{\log }_3}2}}{{1 + {{\log }_3}5}} + \dfrac{1}{{\dfrac{1}{{{{\log }_3}5}} + 1}}\\ = \dfrac{{2{{\log }_3}2}}{{1 + {{\log }_3}5}} + \dfrac{{{{\log }_3}5}}{{{{\log }_3}5 + 1}}\\ = \dfrac{{2{{\log }_3}2 + {{\log }_3}5}}{{{{\log }_3}5 + 1}}\\ = \dfrac{{{{\log }_3}5 + 1 + 2{{\log }_3}2 - 1}}{{{{\log }_3}5 + 1}}\\ = 1 + \dfrac{{2{{\log }_3}2 - 1}}{{{{\log }_3}5 + 1}}\end{array}\)

\( \Rightarrow a = 1,\,\,b =  - 1,\,\,c = 1\).

Vậy \(T = a + b + c = 1 + \left( { - 1} \right) + 1 = 1.\)

Áp dụng công thức ${\log _a}b = \dfrac{1}{{{{\log }_b}a}} \Leftrightarrow {\log _a}b.{\log _b}a = 1\left( {0 < a,b \ne 1} \right)$ ta được:

${\log _2}3 = \dfrac{1}{{{{\log }_3}2}}$ nên D đúng.

Đăt ${\log _2}3 = x$. Ta có

$\begin{array}{l}a = {\log _{12}}18 = \dfrac{{{{\log }_2}18}}{{{{\log }_2}12}} = \dfrac{{{{\log }_2}\left( {{{2.3}^2}} \right)}}{{{{\log }_2}\left( {{2^2}.3} \right)}} = \dfrac{{1 + 2{{\log }_2}3}}{{2 + {{\log }_2}3}} = \dfrac{{1 + 2x}}{{2 + x}}\\ \Rightarrow a\left( {2 + x} \right) = 1 + 2x \Rightarrow x\left( {a - 2} \right) = 1 - 2a\\ \Rightarrow {\log _2}3 = x = \dfrac{{1 - 2a}}{{a - 2}}\end{array}$

Từ công thức ${\log _{{a^n}}}b = \dfrac{1}{n}{\log _a}b\left( {0 < a \ne 1;b > 0;n \ne 0} \right)$ ta thấy chỉ có đáp án B đúng.

$\begin{array}{l}a = {\log _2}14 = {\log _2}2 + {\log _2}7 = 1 + {\log _2}7 \Rightarrow {\log _2}7 = a - 1\\{\log _{49}}32 = {\log _{{7^2}}}{2^5} = \dfrac{5}{2}{\log _7}2 = \dfrac{5}{2}.\dfrac{1}{{{{\log }_2}7}} = \dfrac{5}{{2\left( {a - 1} \right)}}\end{array}$

Ta có: $450 = 150.{e^{5r}}$

$ =  > {e^{5r}} = 3 \Leftrightarrow 5r = \ln 3 =  > r = \dfrac{{\ln 3}}{5}$

Số lượng vi khuẩn sau 10 giờ tăng trưởng là:

$S = 150.{e^{10.\dfrac{{\ln 3}}{5}}} = 150.{\left( {{e^{\ln 3}}} \right)^2} = {150.3^2} = 1350$(con)

Ta có:

${\log _{{a^n}}}b = \dfrac{1}{n}{\log _a}b;{\log _a}\sqrt[n]{b} = \dfrac{1}{n}{\log _a}b$ nên ${\log _{{a^n}}}b = {\log _a}\sqrt[n]{b}$  (C đúng)

Mặt khác: ${\log _{{a^n}}}b = \dfrac{1}{n}{\log _a}b;{\log _{{b^n}}}a = \dfrac{1}{n}{\log _b}a$ nên các đáp án A, B, D đều sai.

Ta có: \({a^2} + 4{b^2} = 5ab \Leftrightarrow {a^2} + 4ab + 4{b^2} = 9ab \Leftrightarrow {\left( {a + 2b} \right)^2} = 9ab\).

Logarit cơ số \(10\) hai vế ta được:

\(\begin{array}{l}\log {\left( {a + 2b} \right)^2} = \log \left( {9ab} \right) \Leftrightarrow 2\log \left( {a + 2b} \right) = \log 9 + \log a + \log b\\ \Leftrightarrow 2\log \left( {a + 2b} \right) = 2\log 3 + \log a + \log b \Leftrightarrow 2\left( {\log \left( {a + 2b} \right) - \log 3} \right) = \log a + \log b\\ \Leftrightarrow \log \dfrac{{a + 2b}}{3} = \dfrac{{\log a + \log b}}{2}\end{array}\)

Ta có: ${\log _{\frac{1}{{\sqrt 3 }}}}81 = {\log _{{{\sqrt 3 }^{ - 1}}}}{3^4} =  - {\log _{\sqrt 3 }}{3^4} $

$=  - {\log _{{3^{\frac{1}{2}}}}}{3^4} =  - \dfrac{1}{{1/2}}{\log _3}{3^4} =  - 2{\log _3}{3^4} =  - 2.4 =  - 8$

$\begin{array}{l}a = {\log _2}60 = {\log _2}\left( {{2^2}.15} \right) = 2 + {\log _2}15 \Rightarrow {\log _2}15 = a - 2\\ \Rightarrow {\log _2}5 = \dfrac{{{{\log }_{15}}5}}{{{{\log }_{15}}2}} = \dfrac{{{{\log }_2}15}}{{{{\log }_5}15}} = \dfrac{{a - 2}}{b}\\b = {\log _5}15 = {\log _5}\left( {3.5} \right) = 1 + {\log _5}3 \Rightarrow {\log _5}3 = b - 1\\{\log _2}3 = {\log _2}5.{\log _5}3 = \dfrac{{a - 2}}{b}.\left( {b - 1} \right) = \dfrac{{ab - 2b - a + 2}}{b}\\{\log _2}12 = {\log _2}\left( {{2^2}.3} \right) = 2 + {\log _2}3 = \dfrac{{ab - a + 2}}{b}\end{array}$

Ta có: ${\log _a}\sqrt {a\sqrt {a\sqrt[3]{a}} }  = {\log _a}\sqrt {a\sqrt {a.{a^{\frac{1}{3}}}} }  = {\log _a}\sqrt {a\sqrt {{a^{\frac{4}{3}}}} }  = {\log _a}\sqrt {a.{a^{\frac{2}{3}}}}  $

$= {\log _a}\sqrt {{a^{\frac{5}{3}}}}  = {\log _a}{a^{\frac{5}{6}}} = \dfrac{5}{6}$

Có $b = {\log _2}6 = 1 + {\log _2}3 \Rightarrow {\log _2}3 = b - 1$

${\log _3}90 = {\log _3}({3^2}.2.5) = 2 + {\log _3}2 + {\log _3}5$ $ = 2 + \dfrac{1}{{{{\log }_2}3}} + \dfrac{{{{\log }_2}5}}{{{{\log }_2}3}} $ $= 2 + \dfrac{{1 + {{\log }_2}5}}{{{{\log }_2}3}} $ $= 2 + \dfrac{{1 + a}}{{b - 1}} = \dfrac{{a + 2b - 1}}{{b - 1}}$

\(\begin{array}{l}{a^2} + 4{b^2} = 12ab \Leftrightarrow {\left( {a + 2b} \right)^2} - 4ab = 12ab \Leftrightarrow {\left( {a + 2b} \right)^2} = 16ab\\ \Rightarrow \ln {\left( {a + 2b} \right)^2} = \ln \left( {16ab} \right)\\ \Rightarrow 2\ln \left( {a + 2b} \right) = \ln 16 + \ln a + \ln b\\ \Rightarrow 2\ln \left( {a + 2b} \right) - 4\ln 2 = \ln a + \ln b\\ \Rightarrow \ln \left( {a + 2b} \right) - 2\ln 2 = \dfrac{1}{2}(\ln a + \ln b)\end{array}\)

Nếu $a > 1$ và $b > c > 0$ thì ${\log _a}b > {\log _a}c$.

Ta có: $\log_a{a^2}{b^4} = \log_a{a^2} + \log_a{b^4} $ $= 2\log_a a + 4\log_a b = 2 + 4p$