Логаритамске једначине 3

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Решити логаритамску једначину.

пр.7)   ${\log _{3x}}\left( {\frac{3}{x}} \right) + \log _3^2x = 1$

пр.8)   ${\log _2}\left( {3 - x} \right) + {\log _2}\left( {1 - x} \right) = 3$

Пр.7

\[\begin{gathered}
{\log _{3x}}\left( {\frac{3}{x}} \right) + \log _3^2x = 1 \hfill \\
3x > 0,3x \ne 1,\frac{3}{x} > 0,x > 0 \Leftrightarrow x > 0,x \ne \frac{1}{3} \hfill \\
\hfill \\
{\log _{3x}}3 - {\log _{3x}}x + \log _3^2x = 1 \hfill \\
\frac{1}{{{{\log }_3}3x}} - \frac{1}{{{{\log }_x}3x}} + \log _3^2x = 1 \hfill \\
\frac{1}{{{{\log }_3}3 + {{\log }_3}x}} - \frac{1}{{{{\log }_x}3 + {{\log }_x}x}} + \log _3^2x = 1 \hfill \\
\frac{1}{{1 + {{\log }_3}x}} - \frac{1}{{\frac{1}{{{{\log }_3}x}} + 1}} + \log _3^2x = 1 \hfill \\
смена:{\log _3}x = t \hfill \\
\frac{1}{{1 + t}} - \frac{1}{{\frac{1}{t} + 1}} + {t^2} = 1 \hfill \\
\frac{1}{{1 + t}} - \frac{1}{{\frac{{1 + t}}{t}}} + {t^2} = 1 \hfill \\
\frac{1}{{1 + t}} - \frac{t}{{1 + t}} + {t^2} = 1\left| { \cdot \left( {1 + t} \right)} \right. \hfill \\
\end{gathered} \]

додатни услов: $1 + t \ne 0 \Leftrightarrow t \ne  - 1$

\[\begin{gathered}
1 - t + {t^2}\left( {1 + t} \right) = 1 + t \hfill \\
1 - t + {t^3} + {t^2} - 1 - t = 0 \hfill \\
{t^3} + {t^2} - 2t = 0 \hfill \\
t\left( {{t^2} + t - 2} \right) = 0 \hfill \\
t = 0,{t^2} + t - 2 = 0 \hfill \\
{t^2} + t - 2 = 0 \hfill \\
{t_{1,2}} = \frac{{ - 1 \pm \sqrt {1 + 8} }}{2} \hfill \\
{t_{1,2}} = \frac{{ - 1 \pm 3}}{2} \hfill \\
{t_1} = - 2,{t_2} = 1 \hfill \\
\hfill \\
t = 0,{t_1} = - 2,{t_2} = 1 \hfill \\
{\log _3}x = 0,{\log _3}x = - 2,{\log _3}x = 1 \hfill \\
{\log _3}x = 0 \hfill \\
x = 1 \hfill \\
\hfill \\
{\log _3}x = - 2 \hfill \\
x = {3^{ - 2}} \hfill \\
x = \frac{1}{9} \hfill \\
\hfill \\
{\log _3}x = 1 \hfill \\
x = 3 \hfill \\
\end{gathered} \]

$x = \left\{ {\frac{1}{9},1,3} \right\}$

Пр.8

\[\begin{gathered}
{\log _2}\left( {3 - x} \right) + {\log _2}\left( {1 - x} \right) = 3 \hfill \\
3 - x > 0;1 - x > 0 \Leftrightarrow x < 1 \hfill \\
{\log _2}\left( {3 - x} \right)\left( {1 - x} \right) = 3 \hfill \\
\left( {3 - x} \right)\left( {1 - x} \right) = {2^3} \hfill \\
3 - 3x - x + {x^2} = 8 \hfill \\
{x^2} - 4x - 5 = 0 \hfill \\
{x_{1,2}} = \frac{{4 \pm \sqrt {16 + 20} }}{2} \hfill \\
{x_{1,2}} = \frac{{4 \pm 6}}{2} \hfill \\
{x_1} = - 1,{x_2} = 5 \hfill \\
\end{gathered} \]

Знамо да je $x < 1.$  Онда је решење jдначине  ${x_1} = -1$