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Bài 70. Giải các phương trình sau:
\(\eqalign{
& a)\,{3^{4x}} = {4^{3x}} \cr
& b)\,{3^{2 – {{\log }_3}x}} = 81x \cr} \)
\(\eqalign{
& c)\,{3^x}{.8^{{x \over {x + 1}}}} = 36 \cr
& d)\,{x^6}{.5^{ – {{\log }_x}5}} = {5^{ – 5}} \cr} \)
\(\eqalign{
& a)\,{3^{4x}} = {4^{3x}} \Leftrightarrow {4^x}{\log _3}3 = {3^x}{\log _3}4 \Leftrightarrow {{{4^x}} \over {{3^x}}} = {\log _3}4 \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Leftrightarrow {\left( {{4 \over 3}} \right)^x} = {\log _3}4 \Leftrightarrow x = {\log _{{4 \over 3}}}\left( {{{\log }_3}4} \right) \cr} \)
Vậy \(S = \left\{ {{{\log }_{{4 \over 3}}}\left( {{{\log }_3}4} \right)} \right\}\)
b) Điều kiện: \(x > 0\)
\(\eqalign{
& {3^{2 – {{\log }_3}x}} = 81x \Leftrightarrow {{{3^2}} \over {{3^{{{\log }_3}x}}}} = 81x \cr
& \Leftrightarrow {9 \over x} = 81x \Leftrightarrow {x^2} = {1 \over 9} \Leftrightarrow x = {1 \over 3}\,\,\left( {\text{ vì }\,x > 0} \right) \cr} \)
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Vậy \(S = \left\{ {{1 \over 3}} \right\}\)
c) Lấy logarit cơ số 3 hai vế ta được:
\(x{\log _3}3 + {x \over {x + 1}}{\log _3}8 = x + {{3x} \over {x + 1}}{\log _3}2 = 2 + 2.{\log _3}2\)
\(\eqalign{
& \Leftrightarrow {x^2} + x + 3\left( {{{\log }_3}2} \right)x = 2x + 2 + 2(x+1)\left( {{{\log }_3}2} \right) \cr
& \Leftrightarrow {x^2} + \left( {{{\log }_3}2 – 1} \right)x – 2.{\log _3}2 -2= 0 \cr
& \Leftrightarrow \left[ \matrix{
x = 2 \hfill \cr
x = – 1 – {\log _3}2 \hfill \cr} \right. \cr} \)
Vậy \(S = \left\{ {2; – 1 – {{\log }_3}2} \right\}\)
d) Điều kiện: \(x > 0\);
Lấy logarit cơ số x hai vế ta được:
\(\eqalign{
& 6 + \left( { – {{\log }_x}5} \right).{\log _x}5 = – 5{\log _x}5 \cr
& \Leftrightarrow \log _x^25 – 5{\log _x}5 – 6 = 0 \cr
& \Leftrightarrow \left[ \matrix{
{\log _x}5 = – 1 \hfill \cr
{\log _x}5 = 6 \hfill \cr} \right. \Leftrightarrow \left[ \matrix{
5 = {x^{ – 1}} \hfill \cr
5 = {x^6} \hfill \cr} \right. \Leftrightarrow \left[ \matrix{
x = {1 \over 5} \hfill \cr
x = \root 6 \of 5 \hfill \cr} \right. \cr} \)
Vậy \(S = \left\{ {{1 \over 5};\root 6 \of 5 } \right\}\)