Video hướng dẫn giải
Giải các phương trình logarit
LG a
a) \({{\log_3}\left( {5x{\rm{ }} + {\rm{ }}3} \right){\rm{ }} = {\rm{ }}{\log_3}\left( {7x{\rm{ }} + {\rm{ }}5} \right)}\)
Phương pháp giải:
+) Tìm điều kiện xác định.
+) Đưa về cùng cơ số: \({\log _a}f\left( x \right) = {\log _a}g\left( x \right) \Leftrightarrow \left\{ \begin{array}{l} f\left( x \right) > 0\\ g\left( x \right) > 0\\ f\left( x \right) = g\left( x \right)\end{array} \right.\)
Lời giải chi tiết:
\(\displaystyle {{\log_3}\left( {5x{\rm{ }} + {\rm{ }}3} \right){\rm{ }} = {\rm{ }}{\log_3}\left( {7x{\rm{ }} + {\rm{ }}5} \right)}\) (1)
\(DK:\left\{ \begin{array}{l}
5x + 3 > 0\\
7x + 5 > 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x > - \dfrac{3}{5}\\
x > - \dfrac{5}{7}
\end{array} \right. \) \(\Leftrightarrow x > - \dfrac{3}{5}\)
TXĐ: \(\displaystyle D = \left( {{{ - 3} \over 5}, + \infty } \right)\)
Khi đó: (1) \(\displaystyle \Rightarrow 5x + 3 = 7x + 5 \) \(\displaystyle ⇔2x=-2 ⇔ x = -1\) (loại)
Vậy phương trình (1) vô nghiệm.
LG b
b) \({\log \left( {x{\rm{ }}-{\rm{ }}1} \right){\rm{ }}-{\rm{ }}\log \left( {2x{\rm{ }}-{\rm{ }}11} \right){\rm{ }} = {\rm{ }}\log {\rm{ }}2}\)
Lời giải chi tiết:
\(\displaystyle {\log\left( {x{\rm{ }}-{\rm{ }}1} \right){\rm{ }}-{\rm{ }}\log\left( {2x{\rm{ }}-{\rm{ }}11} \right){\rm{ }} = {\rm{ }}\log{\rm{ }}2}\) (2)
\(DK:\left\{ \begin{array}{l}
x - 1 > 0\\
2x - 11 > 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x > 1\\
x > \dfrac{{11}}{2}
\end{array} \right. \) \(\Leftrightarrow x > \dfrac{{11}}{2}\)
TXĐ: \(\displaystyle D = \left( {\dfrac{{11}}{2}; + \infty } \right).\)
Khi đó: \(\displaystyle (2) \Rightarrow \log {{x - 1} \over {2x - 11}} = \log 2\) \(\displaystyle \Leftrightarrow {{x - 1} \over {2x - 11}} = 2\) \(\displaystyle \Rightarrow x - 1 = 4x - 22 \Leftrightarrow 3x=21\) \(\displaystyle \Leftrightarrow x = 7 (TM)\)
Vậy phương trình có nghiệm là \(\displaystyle x = 7.\)
LG c
c) \({{\log_2}\left( {x{\rm{ }}-{\rm{ }}5} \right){\rm{ }} + {\rm{ }}{\log_2}\left( {x{\rm{ }} + {\rm{ }}2} \right){\rm{ }} = {\rm{ }}3}\)
Lời giải chi tiết:
\(\displaystyle {{\log_2}\left( {x{\rm{ }}-{\rm{ }}5} \right){\rm{ }} + {\rm{ }}{\log_2}\left( {x{\rm{ }} + {\rm{ }}2} \right){\rm{ }} = {\rm{ }}3}\) (3)
\(DK:\left\{ \begin{array}{l}
x - 5 > 0\\
x + 2 > 0
\end{array} \right. \) \(\Leftrightarrow \left\{ \begin{array}{l}
x > 5\\
x > - 2
\end{array} \right. \Leftrightarrow x > 5\)
TXĐ: \(\displaystyle (5; +∞)\)
Khi đó:
\(\displaystyle (3) \, \Leftrightarrow {\log _2}[(x - 5)(x + 2)]=3\)
\(\displaystyle \Leftrightarrow \left( {x - 5} \right)(x + 2) = 2^3 \)
\( \Leftrightarrow {x^2} - 3x - 10 = 8\)
\(\displaystyle \Leftrightarrow {x^2} - 3x - 18 = 0 \\ \Leftrightarrow (x-6)(x+3)=0 \\ \Leftrightarrow \left[ \matrix{
x - 6=0 \hfill \cr
x + 3=0 \hfill \cr} \right. \Leftrightarrow \left[ \matrix{
x = 6 \, \, (tm) \hfill \cr
x = - 3 \, \,(ktm) \hfill \cr} \right.\)
Vậy phương trình có nghiệm \(\displaystyle x = 6\)
LG d
d) \({\log {\rm{ }}\left( {{x^2}-{\rm{ }}6x{\rm{ }} + {\rm{ }}7} \right){\rm{ }} = {\rm{ }}\log {\rm{ }}\left( {x{\rm{ }}-{\rm{ }}3} \right)}\)
Lời giải chi tiết:
\(\displaystyle {\log_{\rm{ }}\left( {{x^2}-{\rm{ }}6x{\rm{ }} + {\rm{ }}7} \right){\rm{ }} = {\rm{ }}\log{\rm{ }}\left( {x{\rm{ }}-{\rm{ }}3} \right)}\) (4)
\(DK:\left\{ \begin{array}{l}
{x^2} - 6x + 7 > 0\\
x - 3 > 0
\end{array} \right. \) \(\Leftrightarrow \left\{ \begin{array}{l}
\left[ \begin{array}{l}
x > 3 + \sqrt 2 \\
x < 3 - \sqrt 2
\end{array} \right.\\
x > 3
\end{array} \right.\) \( \Leftrightarrow x > 3 + \sqrt 2 \)
TXĐ: \(\displaystyle D = (3 + \sqrt 2 , + \infty )\)
Khi đó:
\(\displaystyle \begin{array}{l}
\left( 4 \right) \Leftrightarrow {x^2} - 6x + 7 = x - 3\\
\Leftrightarrow {x^2} - 7x + 10 = 0\\
\Leftrightarrow \left( {x - 5} \right)\left( {x - 2} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x - 5 = 0\\
x - 2 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 5\;\;\left( {tm} \right)\\
x = 2\;\;\left( {ktm} \right)
\end{array} \right..
\end{array}\)
Vậy phương trình có nghiệm là \(\displaystyle x = 5\).
