Giải bài 38 trang 93 SGK giải tích nâng cao 12

$a)\,\log \dfrac{1}{8}+\dfrac{1}{2}\log 4+4\log \sqrt{2} \\ =\log \dfrac{1}{8}+\log {{4}^{\frac{1}{2}}}+\log {{\left( \sqrt{2} \right)}^{4}} \\ =\log \dfrac{1}{8}+\log 2+\log 4 \\ =\log \dfrac{2.4}{8}=\log 1=0$

$b)\,\log \dfrac{4}{9}+\dfrac{1}{2}\log 36+\dfrac{3}{2}\log \dfrac{9}{2} \\ =\log \dfrac{4}{9}+\log {{36}^{\frac{1}{2}}}+\log {{\left( \dfrac{9}{2} \right)}^{\frac{3}{2}}} \\ =\log \dfrac{4}{9}+\log 6+\log \left( \dfrac{27}{2\sqrt{2}} \right) \\ =\log \dfrac{4.6.27}{9.2\sqrt{2}}=\log \left( 18\sqrt{2} \right)$

$c)\, \log 72-2\log \dfrac{27}{256}+\log \sqrt{108} \\ =\log \left( {{2}^{3}}{{.3}^{2}} \right)-\log \dfrac{{{3}^{6}}}{{{2}^{16}}}+\log \sqrt{{{2}^{{{2}^{3}}}}{{.3}^{3}}} \\ =\log \dfrac{{{2}^{3}}{{.3}^{2}}{{.2}^{16}}.3.2\sqrt{3}}{{{3}^{6}}} \\ =\log \left( {{2}^{20}}{{.3}^{-\frac{5}{2}}} \right) \\ =20\log 2-\dfrac{5}{2}\log 3$

$d)\, \log \dfrac{1}{8}-\log 0,375+2\log \sqrt{0,5625} \\ =\log {{2}^{-3}}-\log \left( 0,{{5}^{3}}.3 \right)+\log \left( 0,{{5}^{4}}{{.3}^{2}} \right) \\ =\log \dfrac{{{2}^{-3}}{{.2}^{-4}}{{.3}^{2}}}{{{2}^{-3}}.3} \\ =\log \left( {{2}^{-4}}.3 \right) \\ =\log 3-4\log 2$

Ghi nhớ: Giả sử a, b, c > 0 và $a\ne 1$. Ta có

$\log_{a}\left( bc \right)={\log }_{a}{b}+{\log }_{a}{c}; \,\,\,\,\,\, {\log _{a}\left( \dfrac{b}{c} \right)={\log }_{a}}{{b}}-{\log }_{a}{c}$

${{\log }_{a}}\left( {{b}^{\alpha }} \right)=\alpha {{\log }_{a}}b;\,\,\,\,\,\,\,{{a}^{{{\log }_{a}}b}}=b;\,\,\,\,\,$