Logarit

${\log _{{a^2}}}\left( {ab} \right) = \dfrac{1}{2}{\log _a}\left( {ab} \right) = \dfrac{1}{2}\left( {{{\log }_a}a + {{\log }_a}b} \right) = \dfrac{1}{2}\left( {1 + {{\log }_a}b} \right) = \dfrac{1}{2} + \dfrac{1}{2}{\log _a}b$

$P = {\log _2}{a^2} - {\log _{{2^{ - 1}}}}{b^2} = {\log _2}{a^2} + {\log _2}{b^2} = {\log _2}\left( {{a^2}{b^2}} \right) = {\log _2}{\left( {ab} \right)^2}$

Điều kiện xác định: $\left\{ \begin{array}{l}x > 0\\{\log _2}x > 0\\{\log _8}x > 0\end{array} \right.$

Khi đó:

${\log _2}\left( {{{\log }_8}x} \right) = {\log _8}\left( {{{\log }_2}x} \right) \Leftrightarrow {\log _2}\left( {\dfrac{1}{3}{{\log }_2}x} \right) = {\log _2}\sqrt[3]{{\left( {{{\log }_2}x} \right)}}$

$\Leftrightarrow \dfrac{1}{3}{\log _2}x = \sqrt[3]{{\left( {{{\log }_2}x} \right)}} \Leftrightarrow \dfrac{1}{{27}}\log _2^3x = {\log _2}x \Leftrightarrow {\left( {{{\log }_2}x} \right)^2} = 27$

(vì ${\log _2}x > 0$ nên chia cả hai vế cho ${\log _2}x \ne 0$

${\log _{{a^2}}}(ab) = \dfrac{1}{2}{\log _a}(ab) = \dfrac{1}{2}(1 + {\log _a}b) = \dfrac{1}{2} + \dfrac{1}{2}{\log _a}b$

Ta có: $\log \left( {a{b^2}} \right) = \log a + \log {b^2} = \log a + 2\log b$

$S = \ln \dfrac{a}{b} + \ln \dfrac{b}{c} + \ln \dfrac{c}{d} + \ln \dfrac{d}{a} = \ln \left( {\dfrac{a}{b}.\dfrac{b}{c}.\dfrac{c}{d}.\dfrac{d}{a}} \right) = \ln 1 = 0$

${\log _{0.5}}a > {\log _{0.5}}b \Leftrightarrow a < b{\rm{  }}$ vì $0,5 <1$ suy ra  A sai.

$\log x < 0 \Leftrightarrow \log x < \log 1 \Leftrightarrow 0 < x < 1$ suy ra B đúng.

${\log _2}x > 0 \Leftrightarrow {\log _2}x > {\log _2}1 \Leftrightarrow x > 1$ suy ra C đúng.

${\log _{\dfrac{1}{3}}}a = {\log _{\dfrac{1}{3}}}b \Leftrightarrow a = b > 0{\rm{  }}$suy ra D đúng.

$\begin{array}{l}P = {\left( {\ln a + {{\log }_a}e} \right)^2} + {\ln ^2}a - \log _a^2e = {\ln ^2}a + 2.\ln a.{\log _a}e + \log _a^2e + {\ln ^2}a - \log _a^2e\\ = 2.{\ln ^2}a + 2.\ln a.\dfrac{{\ln e}}{{\ln a}} = 2{\ln ^2}a + 2\end{array}$

Ta có

$\dfrac{3}{4} < \dfrac{4}{5}$ và ${a^{\frac{3}{4}}} > {a^{\frac{4}{5}}}$$\Rightarrow 0 < a < 1$

$\dfrac{1}{2} < \dfrac{2}{3}$ và ${\log _b}\dfrac{1}{2} < {\log _b}\dfrac{2}{3}$$\Rightarrow b > 1$

Ta có: ${\log _a}b > {\log _a}a = 1;{\log _b}a < {\log _b}b = 1 \Rightarrow {\log _b}a < 1 < {\log _a}b$

Ta có: ${\log _a}\left( {bc} \right) = {\log _a}b + {\log _a}c\left( {0 < a \ne 1;b,c > 0} \right)$

${\log _a}\left( {\dfrac{b}{c}} \right) = {\log _a}b - {\log _a}c\left( {0 < a \ne 1;b,c > 0} \right)$

Vì $0 < x < 1 \Rightarrow \ln x < 0$. Do đó

${\log _c}x > 0 > {\log _b}x > {\log _a}x$ $\Leftrightarrow \dfrac{{\ln x}}{{\ln c}} > 0 > \dfrac{{\ln x}}{{\ln b}} > \dfrac{{\ln x}}{{\ln a}}$ $\Rightarrow \ln c < 0 < \ln a < \ln b$

Mà hàm số $y = \ln x$  đồng biến trên $\left( {0; + \infty } \right)$  nên ta suy ra $c < a < b$

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

Ta có

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

Ta có:

${\log _a}{b^n} = n{\log _a}b\left( {0 < a \ne 1;b > 0} \right)$                                         

${\log _a}\dfrac{1}{b} =  - {\log _a}b\left( {0 < a \ne 1;b > 0} \right)$

${\log _a}\sqrt[n]{b} = {\log _a}{b^{\dfrac{1}{n}}} = \dfrac{1}{n}{\log _a}b\left( {0 < a \ne 1;b > 0;n > 0;n \in {N^*}} \right)$

Vậy đẳng thức không đúng là ${\log _a}\sqrt[n]{b} =  - n{\log _a}b$.

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 ${\log _{10e}}x = \dfrac{{ab}}{{1 + b}}$.

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

Đăt ${\log _2}3 = x$. Ta có

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

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