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Bài 69. Giải các phương trình sau:
\(\eqalign{
& a)\,{\log ^2}{x^3} – 20\log \sqrt x + 1 = 0 \cr
& c)\,{\log _{9x}}27 – {\log _{3x}}243 = 0 \cr} \) \(b)\,{{{{\log }_2}x} \over {{{\log }_4}2x}} = {{{{\log }_8}4x} \over {{{\log }_{16}}8x}}\)
a) Điều kiện: \(x> 0\)
\(\eqalign{
& \,{\log ^2}{x^3} – 20\log \sqrt x + 1 = 0 \Leftrightarrow {\left( {3\log x} \right)^2} – 10\log x + 1 = 0 \cr
& \Leftrightarrow 9{\log ^2}x – 10\log x + 1 = 0 \Leftrightarrow \left[ \matrix{
\log x = 1 \hfill \cr
\log x = {1 \over 9} \hfill \cr} \right. \Leftrightarrow \left[ \matrix{
x = 10 \hfill \cr
x = {10^{{1 \over {9}}}} = \root 9 \of {10} \hfill \cr} \right. \cr} \)
Vậy \(S = \left\{ {10;\root 9 \of {10} } \right\}\)
b) \(\,{{{{\log }_2}x} \over {{{\log }_4}2x}} = {{{{\log }_8}4x} \over {{{\log }_{16}}8x}}\,\,\,\,\,\left( 1 \right)\)
Điều kiện: \(x > 0\), \(x \ne {1 \over 2},\,x \ne {1 \over 8}\)
Ta có: \({\log _4}2x = {{{{\log }_2}2x} \over {{{\log }_2}4}} = {{1 + {{\log }_2}x} \over 2}\)
\(\eqalign{
& {\log _8}4x = {{{{\log }_2}4x} \over {{{\log }_2}8}} = {{2 + {{\log }_2}x} \over 3} \cr
& {\log _{16}}8x = {{{{\log }_2}8x} \over {{{\log }_2}16}} = {{3 + {{\log }_2}x} \over 4} \cr} \)
Đặt \(t = {\log _2}x\) thì (1) thành: \({{2t} \over {1 + t}} = {{4\left( {2 + t} \right)} \over {3\left( {3 + t} \right)}} \Leftrightarrow 6t\left( {3 + t} \right) = 4\left( {1 + t} \right)\left( {2 + t} \right)\)
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\(\eqalign{
& \Leftrightarrow 18t + 6{t^2} = 8 + 12t + 4{t^2} \Leftrightarrow 2{t^2} + 6t – 8 = 0 \Leftrightarrow \left[ \matrix{
t = 1 \hfill \cr
t = – 4 \hfill \cr} \right. \cr
& \left[ \matrix{
{\log _2}x = 1 \hfill \cr
{\log _2}x = – 4 \hfill \cr} \right. \Leftrightarrow \left[ \matrix{
x = 2 \hfill \cr
x = {2^{ – 4}} = {1 \over {16}} \hfill \cr} \right. \cr} \)
Vậy \(S = \left\{ {2;{1 \over {16}}} \right\}\)
c) Điều kiện: \(x > 0\); \(x \ne {1 \over 9},\,x \ne {1 \over 3}\)
Ta có: \({\log _{9x}}27 – {\log _{3x}}3 + {\log _9}243 = 0 \Leftrightarrow {1 \over {{{\log }_{27}}9x}} – {1 \over {{{\log }_3}3x}} + {\log _{{3^2}}}{3^5} = 0\)
\(\eqalign{
& \Leftrightarrow {1 \over {{{\log }_{{3^3}}}9x}} – {1 \over {1 + {{\log }_3}x}} + {5 \over 2} = 0 \cr
& \Leftrightarrow {3 \over {{{\log }_3}9x}} – {1 \over {1 + {{\log }_3}x}} + {5 \over 2} = 0 \cr
& \Leftrightarrow {3 \over {2 + {{\log }_3}x}} – {1 \over {1 + {{\log }_3}x}} + {5 \over 2} = 0 \cr} \)
Đặt \({\log _3}x = t\)
Ta có phương trình: \({3 \over {t + 2}} – {1 \over {t + 1}} + {5 \over 2} = 0\)
\(\eqalign{
& \Leftrightarrow 6\left( {t + 1} \right) – 2\left( {t + 2} \right) + 5\left( {t + 2} \right)\left( {t + 1} \right) = 0 \cr
& \Leftrightarrow \left[ \matrix{
t = – 0,8 \hfill \cr
t = – 3 \hfill \cr} \right. \Leftrightarrow \left[ \matrix{
{\log _3}x = – 0,8 \hfill \cr
{\log _3}x = – 3 \hfill \cr} \right. \Leftrightarrow \left[ \matrix{
x = {3^{ – 0,8}} \hfill \cr
x = {3^{ – 3}} \hfill \cr} \right. \cr} \)
Vậy \(S = \left\{ {{3^{ – 3}};{3^{ – 0,8}}} \right\}\)