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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