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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