Bài 104047

Question

Giải các phương trình logarit:
$a) log^2x – 3logx = log(x^2) – 4$
$b) lo{g_{\frac{1}{3}}}x - 3\sqrt {lo{g_{\frac{1}{3}}}x} + 2 = 0$
$c) lo{g_{\frac{1}{2}}}^2(4x) + lo{g_2}\frac{{{x^2}}}{8} = 8$
$d) lo{g_{\sqrt 5 }}({4^x} - 6) - lo{g_5}{({2^x} - 2)^2} = 2$

score 0 · Answer 1

$a) lg^2 x – 3lgx = lg(x^2) – 4   (*)$
Điều kiện: $x > 0$
$(*) \Leftrightarrow lg^2 x – 3lgx – 2lgx + 4 = 0 \Leftrightarrow lg^2 x – 5lgx + 4 = 0$
Đặt: $u = lgx \Leftrightarrow u^2 – 5u + 4 $ $ \Leftrightarrow $$\left[ \begin{array}{l}
{u_1} = 1\\
{u_2} = 4
\end{array} \right.$
• $ u= 1 \Leftrightarrow lgx = 1 = lg10 \Leftrightarrow x = 10$
•   $ u =4 \Leftrightarrow lgx = 4 = lg10^4 \Leftrightarrow x= 10^4$
$b)$ $lo{g_{\frac{1}{3}}}x - 3\sqrt {lo{g_{\frac{1}{3}}}x} + 2 = 0$
Điều kiện: $\left\{ \begin{array}{l}
x > 0\\
lo{g_{\frac{1}{3}}}x \ge 0 \Leftrightarrow 0 < x \le 1
\end{array} \right.$
Đặt: $u = \sqrt {{{\log }_{\frac{1}{3}}}x} $ = 1 $ \ge $0
($*$)$ \Leftrightarrow u^2 – 3u + 2 = 0$
$u_1 = 1, u_2 = 2$
•    $u = \sqrt {{{\log }_{\frac{1}{3}}}x} $$ \Leftrightarrow $${\log _{\frac{1}{3}}}x$=1$
\Leftrightarrow $x = $\frac{1}{3}$      (nhận)
•    $u = 2 =\sqrt {{{\log }_{\frac{1}{3}}}x} $$ \Leftrightarrow $${\log
_{\frac{1}{3}}}x$=4$ \Leftrightarrow $x = ${\left( {\frac{1}{3}} \right)^4} = \frac{1}{{81}}$   (nhận)
$c$) $lo{g_{\frac{1}{2}}}^2(4x) + lo{g_2}\frac{{{x^2}}}{8} = 8$                 Điều kiện: $x > 0$
$ \Leftrightarrow $$lo{g_{\frac{1}{2}}}^2(4x) + log\frac{{{{\left( {4x} \right)}^2}}}{{8.16}} = 8$
$ \Leftrightarrow $${\left[ { - lo{g_2}(4x)} \right]^2} + log\frac{{{{\left( {4x}
\right)}^2}}}{{28.16}} = 8$
$ \Leftrightarrow $$lo{g_2}^2(4x) + 2lo{g_2}4x = 8 + lo{g_2}({2^6}.2) = 15     (*)     $
Đặt $u = log_2(4x) $
$(*) \Leftrightarrow u^2 + 2u – 15 = 0$
$\Leftrightarrow \left[ \begin{array}{l}
u = 3\\
u = - 5
\end{array} \right.$
•    $u = 3 \Leftrightarrow log_2(4x) = 3 \Leftrightarrow 4x = 2^3 \Leftrightarrow x = 2$
•    $u= -5 \Leftrightarrow log_2(4x) = -5 = log_22^-5 \Leftrightarrow 4x = 2^{-5} \Leftrightarrow x =2^{-7} = \frac{1}{{128}}$
$d$) $lo{g_{\sqrt 5 }}({4^x} - 6) - lo{g_5}{({2^x} - 2)^2} = 2 (*)$
Điều kiện: $\left\{ \begin{array}{l}
{4^x} - 6 > 0\\
{2^x} - 2 > 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
{2^{2x}} - 6 > 0\\
x > 1
\end{array} \right.$
$(*)$ $ \Leftrightarrow $$lo{g_5}{({4^x} - 6)^2} - lo{g_5}{({2^x} - 2)^2} = 2$$ \Leftrightarrow
lo{g_5}{\left( {\frac{{{4^x} - 6}}{{{2^x} - 2}}} \right)^2} = 2$
$ \Leftrightarrow {\mathop{\rm }\nolimits} lo{g_5}{\left( {\frac{{{4^x} - 6}}{{{2^x} - 2}}}
\right)^2} = 1 = {\log _5}5$$ \Leftrightarrow 4^x – 6 = (2^x – 2)5$
$ \Leftrightarrow 2.2^{2x} – 5.2^{2x} + 4 = 0$
Đặt $u = 2^x > 0$
$ \Leftrightarrow u^2 – 5u + 4 \Leftrightarrow $ $\left[ \begin{array}{l}
u = 1 \Leftrightarrow {2^x} = 1 = {2^0} \Leftrightarrow x = 0 {\rm{          }}(loại){\rm{ }}\\
u = 4 \Leftrightarrow {2^x} = 4 = {2^2} \Leftrightarrow x = 2
\end{array} \right.$

Bài 104047

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Bài 104047

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