Giải bài 62 trang 50 sách bài tập toán 11 – Cánh diều

Đề bài

Giải mỗi phương trình sau:

a) \({\log _4}\left( {x – 4} \right) =  – 2;\)

b) \({\log _3}\left( {{x^2} + 2x} \right) = 1;\)

c) \({\log _{25}}\left( {{x^2} – 4} \right) = \frac{1}{2};\)

d) \({\log _9}\left[ {{{\left( {2x – 1} \right)}^2}} \right] = 2;\)

e) \(\log \left( {{x^2} – 2x} \right) = \log \left( {2x – 3} \right);\)

g) \({\log _2}{x^2} + {\log _{\frac{1}{2}}}\left( {2x + 8} \right) = 0.\)

Lời giải chi tiết

a) Điều kiện: \(x > 4.\)

\({\log _4}\left( {x – 4} \right) =  – 2 \Leftrightarrow x – 4 = {4^{ – 2}} \Leftrightarrow x = \frac{{65}}{{16}}\) (thỏa mãn).

b) Điều kiện: \({x^2} + 2x > 0 \Leftrightarrow \left[ \begin{array}{l}x > 0\\x <  – 2\end{array} \right.\)

 \({\log _3}\left( {{x^2} + 2x} \right) = 1 \Leftrightarrow {x^2} + 2x = 3 \Leftrightarrow {x^2} + 2x – 3 = 0 \Leftrightarrow \left[ \begin{array}{l}x = 1\\x =  – 3\end{array}  \right)\) (thỏa mãn)

c) Điều kiện: \({x^2} – 4 > 0 \Leftrightarrow \left[ \begin{array}{l}x > 2\\x <  – 2\end{array} \right.\)

\({\log _{25}}\left( {{x^2} – 4} \right) = \frac{1}{2} \Leftrightarrow {x^2} – 4 = {25^{\frac{1}{2}}} \Leftrightarrow {x^2} – 4 = 5 \Leftrightarrow \left[ \begin{array}{l}x = 3\\x =  – 3\end{array}  \right)\) (thỏa mãn)

d) Điều kiện: \({\left( {2x – 1} \right)^2} > 0 \Leftrightarrow x \ne \frac{1}{2}.\)

\({\log _9}\left[ {{{\left( {2x – 1} \right)}^2}} \right] = 2 \Leftrightarrow {\left( {2x – 1} \right)^2} = {9^2} \Leftrightarrow \left[ \begin{array}{l}2x – 1 = 9\\2x – 1 =  – 9\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = 5\\x =  – 4\end{array}  \right)\) (thỏa mãn)

e) \(\log \left( {{x^2} – 2x} \right) = \log \left( {2x – 3} \right) \Leftrightarrow \left\{ \begin{array}{l}{x^2} – 2x = 2x – 3\\2x – 3 > 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}{x^2} – 4x + 3 = 0\\x > \frac{3}{2}\end{array} \right.\)

\( \Leftrightarrow \left\{ \begin{array}{l}\left[ \begin{array}{l}x = 1\\x = 3\end{array} \right.\\x > \frac{3}{2}\end{array} \right. \Leftrightarrow x = 3.\)

g) Điều kiện: \(\left\{ \begin{array}{l}{x^2} > 0\\2x + 8 > 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x >  – 4\\x \ne 0\end{array} \right..\)

 \(\begin{array}{l}{\log _2}{x^2} + {\log _{\frac{1}{2}}}\left( {2x + 8} \right) = 0 \Leftrightarrow {\log _2}{x^2} – {\log _2}\left( {2x + 8} \right) = 0 \Leftrightarrow {\log _2}\frac{{{x^2}}}{{2x + 8}} = 0\\ \Leftrightarrow \frac{{{x^2}}}{{2x + 8}} = 1 \Leftrightarrow {x^2} = 2x + 8 \Leftrightarrow {x^2} – 2x – 8 = 0 \Leftrightarrow \left[ \begin{array}{l}x = 4\\x =  – 2\end{array} \right.\left( {TM} \right).\end{array}\)