Giải mỗi phương trình sau:
a) \(0,{5^{2{x^2} + x - 1}} = \frac{1}{4};\)
b) \({2^{{x^2} - 6x - \frac{5}{2}}} = 16\sqrt 2 ;\)
c) \({27^{{x^2} - 4x + 4}} = {9^{{x^2} - 4}};\)
d) \(0,{05^{x - 3}} = {\left( {2\sqrt 5 } \right)^x};\)
e) \({\log _3}3\left( {x - 2} \right) = - 1;\)
g) \({\log _5}\left( {{x^2} + 1} \right) + {\log _{0,2}}\left( {4 - 5x - {x^2}} \right) = 0.\)
Giải phương trình bằng định nghĩa hàm số lôgarit hoặc đưa về cùng cơ số kết hợp biến đổi sử dụng công thức lôgarit.
Advertisements (Quảng cáo)
a) \(0,{5^{2{x^2} + x - 1}} = \frac{1}{4} \Leftrightarrow 0,{5^{2{x^2} + x - 1}} = 0,{5^2} \Leftrightarrow 2{x^2} + x - 1 = 2 \Leftrightarrow 2{x^2} + x - 3 = 0\)
\( \Leftrightarrow \left[ \begin{array}{l}x = 1\\x = - \frac{3}{2}\end{array} \right.\)
b) \({2^{{x^2} - 6x - \frac{5}{2}}} = 16\sqrt 2 \Leftrightarrow {2^{{x^2} - 6x - \frac{5}{2}}} = {2^4}{.2^{\frac{1}{2}}} \Leftrightarrow {2^{{x^2} - 6x - \frac{5}{2}}} = {2^{\frac{9}{2}}} \Leftrightarrow {x^2} - 6x - \frac{5}{2} = \frac{9}{2}\)
\( \Leftrightarrow {x^2} - 6x - 7 = 0 \Leftrightarrow \left[ \begin{array}{l}x = - 1\\x = 7\end{array} \right.\)
c) \({27^{{x^2} - 4x + 4}} = {9^{{x^2} - 4}} \Leftrightarrow {3^{3\left( {{x^2} - 4x + 4} \right)}} = {3^{2\left( {{x^2} - 4} \right)}} \Leftrightarrow 3\left( {{x^2} - 4x + 4} \right) = 2\left( {{x^2} - 4} \right)\)
\( \Leftrightarrow {x^2} - 12x + 20 = 0 \Leftrightarrow \left[ \begin{array}{l}x = 10\\x = 2\end{array} \right.\)
d) \(0,{05^{x - 3}} = {\left( {2\sqrt 5 } \right)^x} \Leftrightarrow {\left( {\frac{1}{{20}}} \right)^{x - 3}} = {\left( {2\sqrt 5 } \right)^x} \Leftrightarrow {\left( {{{\left( {2\sqrt 5 } \right)}^2}} \right)^{3 - x}} = {\left( {2\sqrt 5 } \right)^x}\)
\( \Leftrightarrow {\left( {2\sqrt 5 } \right)^{2\left( {3 - x} \right)}} = {\left( {2\sqrt 5 } \right)^x} \Leftrightarrow 6 - 2x = x \Leftrightarrow x = 2.\)
e) \({\log _3}3\left( {x - 2} \right) = - 1 \Leftrightarrow 3\left( {x - 2} \right) = \frac{1}{3} \Leftrightarrow x - 2 = \frac{1}{9} \Leftrightarrow x = \frac{{19}}{9}.\)
g) \({\log _5}\left( {{x^2} + 1} \right) + {\log _{0,2}}\left( {4 - 5x - {x^2}} \right) = 0 \Leftrightarrow {\log _5}\left( {{x^2} + 1} \right) + {\log _{{5^{ - 1}}}}\left( {4 - 5x - {x^2}} \right) = 0\)
\(\begin{array}{l} \Leftrightarrow {\log _5}\left( {{x^2} + 1} \right) = {\log _5}\left( {4 - 5x - {x^2}} \right) \Leftrightarrow \left\{ \begin{array}{l}{x^2} + 1 = 4 - 5x - {x^2}\\{x^2} + 1 > 0\end{array} \right.\\ \Leftrightarrow 2{x^2} + 5x - 3 = 0 \Leftrightarrow \left[ \begin{array}{l}x = - 3\\x = \frac{1}{2}\end{array} \right.\end{array}\)