Bài 100811

Question

Giải các phương trình :
$\begin{array}{l}
1) & {\log ^2}{x^3} - 20\log \sqrt x + 1 = 0\\
2) & {\log _{\sqrt x }}2 + 4{\log _4}{x^2} + 9 = 0\\
3) & {\log _x}3.{\log _{\frac{x}{3}}}3 + {\log _{\frac{x}{{81}}}}3 = 0\\
4) & {x^{{{\left( {{{\log }_3}x} \right)}^3} - 3{{\log }_3}x}} = {3^{8 - {{\log }_{\sqrt 2 }}4}}
\end{array}$

score 0 · Answer 1

1)
Điều kiện: $x>0.$
Khi đó phương trình đã cho trở thành
$(3\log x)^2-20.\frac{1}{2}\log x+1=0$
$\Leftrightarrow 9\log^2x-10\log x+1=0$
$\Leftrightarrow (9\log x-1)(\log x-1)=0$
$\Leftrightarrow \left[ \begin{array}{l} \log x=\frac{1}{9}\\\log x=1 \end{array} \right.\Leftrightarrow \left[\begin{array}{l} x=10^\frac{1}{9}=\sqrt[9]{10}\\ x=10^1=10. \end{array} \right.(TM)$

2)
Điều kiện: $x>0 , x\neq 1.$
Khi đó phương trình đã cho trở thành
$\log_{x^\frac{1}{2}}2+4.\log_{2^2}x^2+9=0$
$\Leftrightarrow 2\log_x2+\frac{4}{\log_x2}+9=0$
$\Leftrightarrow 2\log_x^22+9\log_x2+4=0$
$\Leftrightarrow (2\log_x2+1)(\log_x2+4)=0$
$\Leftrightarrow \left[ \begin{array}{l} \log_2 x=-2\\ \log_2x=-\frac{1}{4} \end{array} \right.\Leftrightarrow \left[\begin{array}{l} x=2^{-2}=\frac{1}{4}\\ x=2^{-\frac{1}{4}}=\frac{1}{{\sqrt[4]{2}}}. \end{array} \right.$

3)
Điều kiện: $x>0 ,x\neq 3 ,x\neq 81.$
Khi đó phương trình đã cho trở thành
$\frac{1}{\log_3x}.\frac{1}{\log_3\frac{x}{3}}+\frac{1}{\log_3\frac{x}{81}}=0$
$\Leftrightarrow (\log_3x-\log_381)+\log_3x(\log_3x-\log_33)=0$
$\Leftrightarrow \log^2_3x-4=0$
$\Leftrightarrow (\log_3x+2)(\log_3x-2)=0$
$\Leftrightarrow \left[ \begin{array}{l} \log_3x=-2\\\log_3x=2 \end{array} \right.\Leftrightarrow \left[ \begin{array}{l} x=3^{-2}=\frac{1}{9}\\ x=3^2=9 \end{array} \right.(TM)$

4)
Điều kiện: $x>0.$
Do 2 vế PT đều dương, lấy logarit cơ số 3 của 2 vế, khi đó phương trình đã cho trở thành
$\log_3(x^{\log^3_3x-3\log_3x})=\log_33^{8-\log_{\sqrt2}4}$
$\Leftrightarrow (\log^3_3x-3\log_3x)\log_3x=(8-\log_{2^\frac{1}{2}}2^2)\log_33$
$\Leftrightarrow \log^4_3x-3\log^2_3x-8+4=0$
$\Leftrightarrow (\log^2_3x-4)(\log^2_3x+1)=0$
Do $\log_3^2x\geq 0\forall x\Rightarrow \log_3^2x=4$
$\Rightarrow \log_3x=\pm 2\Leftrightarrow \left[\begin{array}{l} x=3^2=9\\x=3^{-2}=\frac{1}{9} \end{array} \right.(TM)$.

ĐS :
$\begin{array}{l}
1) & x = 10;\,x = \sqrt[9]{{10}} & & 3) & x = 9;\,x = \frac{1}{9}\\
2) & x = \frac{1}{4};\,x = \frac{1}{{\sqrt[4]{2}}} & & 4) & x = 9;\,x = \frac{1}{9}
\end{array}$

Bài 100811

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Bài 100811

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