Các bài toán thường gặp về logarit

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

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

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

Ta có $80 = {4^2}.5;{\rm{    }}12 = 3.4$

$\begin{array}{l}{\log _{12}}80 = {\log _{12}}{4^2} + {\log _{12}}5 = 2{\log _{12}}4 + {\log _{12}}5 = \dfrac{2}{{{{\log }_4}12}} + \dfrac{1}{{{{\log }_5}12}} = \dfrac{2}{{{{\log }_4}3 + 1}} + \dfrac{1}{{{{\log }_5}3 + {{\log }_5}4}}\\ = \dfrac{2}{{\dfrac{1}{a} + 1}} + \dfrac{1}{{\dfrac{b}{a} + b}} = \dfrac{{2a}}{{a + 1}} + \dfrac{a}{{b\left( {a + 1} \right)}} = \dfrac{{2ab + a}}{{ab + b}}\end{array}$

Gán giá trị đề bài cho bằng cách bấm:

Lần lượt thử từng đáp án:

Ta có ${\log _{{a^2}}}b + {\log _{{b^2}}}a = 1 \Leftrightarrow \dfrac{1}{2}({\log _a}b + {\log _b}a) = 1 \Leftrightarrow {\log _a}b + {\log _b}a = 2$

Vì ${\log _a}b.{\log _b}a = 1$  nên ${\log _a}b,{\log _b}a$ là nghiệm của phương trình${x^2} - 2x + 1 = 0$. Suy ra  ${\log _a}b = {\log _b}a = 1$  hay $a = b$.

Dựa vào 2 kết quả trên ta có

$\begin{array}{l}m = \left[ {\log {2^{30}}} \right] + 1 = \left[ {30\log 2} \right] + 1 = 10\\n = \left[ {{{\log }_2}{{30}^2}} \right] + 1 = \left[ {2{{\log }_2}30} \right] + 1 = 10\\ \Rightarrow m + n = 20\end{array}$

Ta có:

$\begin{array}{l}\,\,\,{\log _{12}}35 = {\log _{12}}\left( {5.7} \right)\\ = {\log _{12}}5 + {\log _{12}}7\\ = \dfrac{1}{{{{\log }_5}\left( {{2^2}.3} \right)}} + \dfrac{1}{{{{\log }_7}\left( {{2^2}.3} \right)}}\\ = \dfrac{1}{{2{{\log }_5}2 + {{\log }_5}3}} + \dfrac{1}{{2{{\log }_7}2 + {{\log }_7}3}}\end{array}$

Theo bài ra ta có:

$\begin{array}{l}a = {\log _{27}}5 = \dfrac{1}{{{{\log }_5}{3^3}}} = \dfrac{1}{{3{{\log }_5}3}}\\ \Rightarrow {\log _5}3 = \dfrac{1}{{3a}}\\b = {\log _8}7 = \dfrac{1}{{{{\log }_7}{2^3}}} = \dfrac{1}{{3{{\log }_7}2}}\\ \Rightarrow {\log _7}2 = \dfrac{1}{{3b}}\\c = {\log _2}3 \Rightarrow {\log _3}2 = \dfrac{1}{c}\\ \Rightarrow {\log _5}2 = {\log _5}3.{\log _3}2 = \dfrac{1}{{3a}}.\dfrac{1}{c} = \dfrac{1}{{3ac}}\\\,\,\,\,\,\,{\log _7}3 = {\log _7}2.{\log _2}3 = \dfrac{1}{{3b}}.c = \dfrac{c}{{3b}}\end{array}$

Khi đó ta có:

$\begin{array}{l}\,\,\,{\log _{12}}35 = {\log _{12}}\left( {5.7} \right)\end{array}$

$= \dfrac{1}{{\dfrac{2}{{3ac}} + \dfrac{1}{{3a}}}} + \dfrac{1}{{\dfrac{2}{{3b}} + \dfrac{c}{{3b}}}}$

$= \dfrac{{3ac}}{{2 + c}} + \dfrac{{3b}}{{2 + c}} = \dfrac{{3ac + 3b}}{{c + 2}}$

${\log _2}a + 3{\log _2}b = {\log _2}a + {\log _2}{b^3} = {\log _2}\left( {a{b^3}} \right) = {\log _2}8 = 3.$