Giải bài 70 trang 125 SGK giải tích nâng cao 12

Gợi ý: Sử dụng phương pháp lôgarit hóa

$\begin{align} a)\,& {{3}^{{{4}^{x}}}}={{4}^{3}}^{^{x}} \\ & {{\log }_{3}}{{3}^{{{4}^{x}}}}={{\log }_{3}}{{4}^{{{3}^{x}}}} \\ & \Leftrightarrow {{4}^{x}}={{3}^{x}}{{\log }_{3}}4 \\ & \Leftrightarrow {{\left( \dfrac{4}{3} \right)}^{x}}={{\log }_{3}}4 \\ & \Leftrightarrow x={{\log }_{\frac{4}{3}}}\left( {{\log }_{3}}4 \right) \\ b)\,& {{3}^{2-{{\log }_{3}}x}}=81x \\ & \Leftrightarrow {{\log }_{3}}{{3}^{2-{{\log }_{3}}x}}={{\log }_{3}}81x \\ & \Leftrightarrow 2-{{\log }_{3}}x=4+{{\log }_{3}}x \\ & \Leftrightarrow 2{{\log }_{3}}x=-2 \\ & \Leftrightarrow x=\dfrac{1}{3} \end{align}$

$c)\,{{3}^{x}}{{.8}^{\dfrac{x}{x+1}}}=36 \\ \Leftrightarrow {{\log }_{3}}\left( {{3}^{x}}{{.2}^{\dfrac{3x}{x+1}}} \right)={{\log }_{3}}\left( {{3}^{2}}{{.2}^{2}} \right) \\ \Leftrightarrow x+\dfrac{3x}{x+1}{{\log }_{3}}2=2+2{{\log }_{3}}2 \\ \Leftrightarrow {{x}^{2}}+x+3x{{\log }_{3}}2=\left( 2+2{{\log }_{3}}2 \right)x+2+2{{\log }_{3}}2 \\ \Leftrightarrow {{x}^{2}}+\left( {{\log }_{3}}2-1 \right)x-2-2{{\log }_{3}}2=0$

Ta có 

$\Delta =\log _{3}^{2}2-2{{\log }_{3}}2+1+8+8{{\log }_{3}}2 \\ \,\,\,\, ={{\left( {{\log }_{3}}2+3 \right)}^{2}} \\ \Leftrightarrow \left[ \begin{align} & x=\dfrac{1-{{\log }_{3}}2+{{\log }_{3}}2+3}{2}=2 \\ & x=\dfrac{1-{{\log }_{3}}2-{{\log }_{3}}2-3}{2}=-1-{{\log }_{3}}2 \\ \end{align} \right.$

$\begin{aligned} d)\, & {{x}^{6}}{{.5}^{-{{\log }_{x}}5}}={{5}^{-5}} \\ & \Leftrightarrow {{\log }_{x}}\left( {{x}^{6}}{{.5}^{-{{\log }_{x}}5}} \right)={{\log }_{x}}{{5}^{-5}} \\ & \Leftrightarrow 6-{{\log }_{x}}5.{{\log }_{x}}5=-5{{\log }_{x}}5 \\ & \Leftrightarrow {{\left( {{\log }_{5}}x \right)}^{2}}-5{{\log }_{x}}5-6=0 \\ & \Leftrightarrow \left[ \begin{aligned} & {{\log }_{x}}5=-1 \\ & {{\log }_{x}}5=6 \\ \end{aligned} \right. \\ & \Leftrightarrow \left[ \begin{aligned} & 5=\dfrac{1}{x} \\ & 5={{x}^{6}} \\ \end{aligned} \right. \\ & \Leftrightarrow \left[ \begin{aligned} & x=\dfrac{1}{5} \\ & x=\sqrt[6]{5} \\ \end{aligned} \right. \\ \end{aligned}$