Giải bài 75 trang 127 SGK giải tích nâng cao 12

a) Điều kiện: $x>0$.

${{\log }_{3}}\left( {{3}^{x}}-1 \right).{{\log }_{3}}\left( {{3}^{x+1}}-3 \right)=12 \\ {{\log }_{3}}\left( {{3}^{x}}-1 \right)\left[ 1+{{\log }_{3}}\left( {{3}^{x}}-1 \right) \right]=12 \\ \Leftrightarrow \log _{3}^{2}\left( {{3}^{x}}-1 \right)+{{\log }_{3}}\left( {{3}^{x}}-1 \right)-12=0 \\ \Leftrightarrow \left[ \begin{aligned} & {{\log }_{3}}\left( {{3}^{x}}-1 \right)=3 \\ & {{\log }_{3}}\left( {{3}^{x}}-1 \right)=-4 \\ \end{aligned} \right. \\ \Leftrightarrow \left[ \begin{aligned} & {{3}^{x}}-1=27 \\ & {{3}^{x}}-1=\dfrac{1}{81} \\ \end{aligned} \right. \\ \Leftrightarrow \left[ \begin{aligned} & {{3}^{x}}=28 \\ & {{3}^{x}}=\dfrac{82}{81} \\ \end{aligned} \right.\Leftrightarrow \left[ \begin{aligned} & x={{\log }_{3}}28 \\ & x={{\log }_{3}}\dfrac{82}{81} \\ \end{aligned} \right.$

b) Điều kiện: $x>1,x\ne 2$.

${{\log }_{x-1}}4=1+{{\log }_{2}}\left( x-1 \right) \\ \Leftrightarrow 2{{\log }_{x-1}}2=1+\dfrac{1}{{{\log }_{x-1}}2} \\ \Leftrightarrow 2{{\left( {{\log }_{x-1}}2 \right)}^{2}}+{{\log }_{x-1}}2-1=0 \\ \Leftrightarrow \left[ \begin{aligned} & {{\log }_{x-1}}2=-1 \\ & {{\log }_{x-1}}2=\dfrac{1}{2} \\ \end{aligned} \right.\Leftrightarrow \left[ \begin{aligned} & 2=\dfrac{1}{x-1} \\ & 2=\sqrt{x-1} \\ \end{aligned} \right. \\ \Leftrightarrow \left[ \begin{aligned} & x-1=\dfrac{1}{2} \\ & x-1=4 \\ \end{aligned} \right.\Leftrightarrow \left[ \begin{aligned} & x=\dfrac{3}{2} \\ & x=5 \\ \end{aligned} \right.$

c) Điều kiện: $\left\{ \begin{aligned} & {{\log }_{2}}\left( -x \right)\ge 0 \\ & {{x}^{2}}>0 \\ & -x>0 \\ \end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned} & -x\ge 1 \\ & x\ne 0 \\ & x<0 \\ \end{aligned} \right.\Leftrightarrow x\le -1$

$5\sqrt{{{\log }_{2}}\left( -x \right)}={{\log }_{2}}\sqrt{{{x}^{2}}} \\ \Leftrightarrow 5\sqrt{{{\log }_{2}}\left( -x \right)}={{\log }_{2}}\left( -x \right) \\ \Leftrightarrow \sqrt{{{\log }_{2}}\left( -x \right)}\left( 5-\sqrt{{{\log }_{2}}\left( -x \right)} \right)=0 \\ \Leftrightarrow \left[ \begin{aligned} & {{\log }_{2}}\left( -x \right)=0 \\ & {{\log }_{2}}\left( -x \right)=25 \\ \end{aligned} \right. \\ \Leftrightarrow \left[ \begin{aligned} & -x=1 \\ & -x={{2}^{25}} \\ \end{aligned} \right.\Leftrightarrow \left[ \begin{aligned} & x=-1 \\ & x=-{{2}^{25}} \\ \end{aligned} \right.$

d) Điều kiện: $x>0$.

${{3}^{{{\log }_{4}}x+\frac{1}{2}}}+{{3}^{{{\log }_{4}}x-\frac{1}{2}}}=\sqrt{x} \\ \Leftrightarrow \sqrt{3}{{.3}^{{{\log }_{4}}x}}+\dfrac{1}{\sqrt{3}}{{.3}^{{{\log }_{4}}x}}=\sqrt{x} \\ \Leftrightarrow \dfrac{4}{\sqrt{3}}{{.3}^{{{\log }_{4}}x}}={{2}^{{{\log }_{2}}\sqrt{x}}}={{2}^{{{\log }_{4}}x}} \\ \Leftrightarrow {{\left( \dfrac{3}{2} \right)}^{{{\log }_{4}}x}}=\dfrac{\sqrt{3}}{4} \\ \Leftrightarrow {{\log }_{4}}x={{\log }_{\frac{3}{2}}}\dfrac{\sqrt{3}}{4} \\ \Leftrightarrow x={{4}^{{{\log }_{\frac{3}{2}}}\frac{\sqrt{3}}{4}}}$