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Câu hỏi: Cho ${{\log }_{2}}\left[ {{\log }_{\dfrac{1}{2}}}\left( {{\log }_{2}}x \right) \right]={{\log }_{3}}\left[ {{\log }_{\dfrac{1}{3}}}\left( {{\log }_{3}}y \right) \right]={{\log }_{5}}\left[ {{\log }_{\dfrac{1}{5}}}\left( {{\log }_{5}}z \right) \right]=0.$ Khẳng định nào sau đây đúng?
A. $y<z<x$
B. $x<y<z$
C. $z<x<y$
D. $z<y<x$
A. $y<z<x$
B. $x<y<z$
C. $z<x<y$
D. $z<y<x$
Phương pháp:
Giải phương trình logarit: ${{\log }_{a}}f\left( x \right)=b\Leftrightarrow f\left( x \right)={{a}^{b}}$ tìm $x,y,z$ và so sánh.
Cách giải:
ĐKXĐ: $\left\{ \begin{aligned}
& x,y,z>0 \\
& {{\log }_{2}}x>0 \\
& {{\log }_{3}}y>0 \\
& {{\log }_{5}}z>0 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& x>1 \\
& y>1 \\
& z>1 \\
\end{aligned} \right..$
Ta có:
${{\log }_{2}}\left[ {{\log }_{\dfrac{1}{2}}}\left( {{\log }_{2}}x \right) \right]={{\log }_{3}}\left[ {{\log }_{\dfrac{1}{3}}}\left( {{\log }_{3}}y \right) \right]={{\log }_{5}}\left[ {{\log }_{\dfrac{1}{5}}}\left( {{\log }_{5}}z \right) \right]=0$
$\Leftrightarrow {{\log }_{\dfrac{1}{2}}}\left( {{\log }_{2}}x \right)={{\log }_{\dfrac{1}{3}}}\left( {{\log }_{3}}y \right)={{\log }_{\dfrac{1}{5}}}\left( {{\log }_{5}}z \right)=1$
$\Leftrightarrow \left\{ \begin{aligned}
& {{\log }_{2}}x=\dfrac{1}{2} \\
& {{\log }_{3}}y=\dfrac{1}{3} \\
& {{\log }_{5}}z=\dfrac{1}{5} \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& x={{2}^{\dfrac{1}{2}}}\approx 1,41 \\
& y={{3}^{\dfrac{1}{3}}}\approx 1,44 \\
& z={{5}^{\dfrac{1}{5}}}\approx 1,37 \\
\end{aligned} \right.$
Vậy $z<x<y.$
Giải phương trình logarit: ${{\log }_{a}}f\left( x \right)=b\Leftrightarrow f\left( x \right)={{a}^{b}}$ tìm $x,y,z$ và so sánh.
Cách giải:
ĐKXĐ: $\left\{ \begin{aligned}
& x,y,z>0 \\
& {{\log }_{2}}x>0 \\
& {{\log }_{3}}y>0 \\
& {{\log }_{5}}z>0 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& x>1 \\
& y>1 \\
& z>1 \\
\end{aligned} \right..$
Ta có:
${{\log }_{2}}\left[ {{\log }_{\dfrac{1}{2}}}\left( {{\log }_{2}}x \right) \right]={{\log }_{3}}\left[ {{\log }_{\dfrac{1}{3}}}\left( {{\log }_{3}}y \right) \right]={{\log }_{5}}\left[ {{\log }_{\dfrac{1}{5}}}\left( {{\log }_{5}}z \right) \right]=0$
$\Leftrightarrow {{\log }_{\dfrac{1}{2}}}\left( {{\log }_{2}}x \right)={{\log }_{\dfrac{1}{3}}}\left( {{\log }_{3}}y \right)={{\log }_{\dfrac{1}{5}}}\left( {{\log }_{5}}z \right)=1$
$\Leftrightarrow \left\{ \begin{aligned}
& {{\log }_{2}}x=\dfrac{1}{2} \\
& {{\log }_{3}}y=\dfrac{1}{3} \\
& {{\log }_{5}}z=\dfrac{1}{5} \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& x={{2}^{\dfrac{1}{2}}}\approx 1,41 \\
& y={{3}^{\dfrac{1}{3}}}\approx 1,44 \\
& z={{5}^{\dfrac{1}{5}}}\approx 1,37 \\
\end{aligned} \right.$
Vậy $z<x<y.$
Đáp án C.