Bài 104046

Question

Giải các phương trình logarit :

$a) log_4(x+3) – log_4(x-1) = 2 – log_48$

$b) log_{5-x}(x^2-2x + 65) = 2$

$c) log5 + log(x + 10) – 1 = log(21x – 20) – log(2x – 1)$

$d) logx- \frac{1}{2}log(x - \frac{1}{2}) = log(x + \frac{1}{2}) - \frac{1}{2}log(x + \frac{1}{8})$

score 0 · Answer 1

$a) log_4(x+3) – log_4(x-1) = 2 – log_48$
Điều kiện :

$\begin{array}{l} \left\{ \begin{array}{l} x + 3 > 0\\ x - 1 > 0 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} x > - 3\\ x > 1 \end{array} \right. \Leftrightarrow x > 1\\ \Leftrightarrow lo{g_4}(x + 3) = lo{g_4}{4^2} - lo{g_4}8 = lo{g_4}\frac{{16}}{8}\\ \Leftrightarrow \frac{{x + 3}}{{x - 1}} = 2 \Leftrightarrow x = 5 > 1\\ \end{array}$

$b) log_{5-x}(x^2-2x + 65) = 2$ , vì :

$x^2-2x + 65>0$ (

$\delta = 1- 65 = - 64<0)$

$\Leftrightarrow \left\{ \begin{array}{l} lo{g_{5 - x}}\left( {{x^2} - 2x + 65} \right) = lo{g_{5 - x}}{(5 - x)^2}\\ 5 - x > 0,x\neq 4 \end{array} \right.$

$\Leftrightarrow \left\{ \begin{array}{l} {x^2} - 2x + 65 = {(5 - x)^2}\\ x < 5,x\neq 4 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} 8x = - 40\\ x < 5 va x \neq 4 \end{array} \right. \Rightarrow x = - 5$

$c) log5 + log(x + 10) – 1 = log(21x – 20) – log(2x – 1)$
Điều kiện:

$\begin{array}{l} \left\{ \begin{array}{l} x + 10 > 0\\ 21x - 20 > 0\\ 2x - 1 > 0 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} x > - 10\\ x > \frac{{20}}{{21}}\\ x > \frac{1}{2} \end{array} \right. \Leftrightarrow x > \frac{{20}}{{21}}\\ \end{array}$

$\Leftrightarrow log5(x + 10) - log10 = log\frac{{21x - 20}}{{2x - 1}}$

$\Leftrightarrow log\frac{{5(x + 10)}}{{10}} = log\frac{{21x - 20}}{{2x - 1}} \Leftrightarrow \frac{{(x + 10)}}{2} = \frac{{21x - 20}}{{2x - 1}}$

$\Leftrightarrow 2{x^2} - 23x + 30 = 0$

$V = 289 = 1{7^2} \Leftrightarrow \left[ \begin{array}{l} {x_1} = \frac{{23 + 17}}{4} = 10 (nhận)\\ {x_2} = \frac{{23 - 17}}{4} = \frac{3}{2} (nhận) \end{array} \right.$

$d) logx \frac{1}{2}log(x - \frac{1}{2}) = log(x + \frac{1}{2}) - \frac{1}{2}log(x + \frac{1}{8})$
Điều kiện:

$\left\{ \begin{array}{l} x > 0\\ x > \frac{1}{2}\\ x > - \frac{1}{2}\\ x > - \frac{1}{8} \end{array} \right. \Rightarrow x > \frac{1}{2}$
Tương tự như câu

$c$ ) ta có:

$log\frac{x}{{\sqrt {x - 1} }} = log\frac{{x + \frac{1}{2}}}{{\sqrt {x + \frac{1}{8}} }}$

$\Leftrightarrow \frac{{{x^2}}}{{x - \frac{1}{2}}} = \frac{{{{\left( {x + \frac{1}{2}} \right)}^2}}}{{x + \frac{1}{8}}} \Leftrightarrow \left[ \begin{array}{l} x = 1 (nhận)\\ x = \frac{1}{3} ( loại ) \end{array} \right.$

Bài 104046

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Bài 104046

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