Логаритамске једначине 4

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Решити логаритамску једначину.

пр.9)    $\lg 2 + \lg \left( {{4^{x - 2}} + 9} \right) = 1 + \lg \left( {{2^{x - 2}} + 1} \right)$

пр.10)  $\log _{\frac{1}{2}}^24x + {\log _2}\left( {\frac{{{x^2}}}{8}} \right) = 8$

Пр.9

\[\begin{gathered}
\lg 2 + \lg \left( {{4^{x - 2}} + 9} \right) = 1 + \lg \left( {{2^{x - 2}} + 1} \right) \hfill \\
{4^{x - 2}} + 9 > 0 \hfill \\
{2^{x - 2}} + 1 > 0 \hfill \\
\end{gathered} \]

Ови услови важе $\forall x \in R$

\[\begin{gathered}
\lg 2\left( {{4^{x - 2}} + 9} \right) = \lg 10 + \lg \left( {{2^{x - 2}} + 1} \right) \hfill \\
\lg 2\left( {{4^{x - 2}} + 9} \right) = \lg 10\left( {{2^{x - 2}} + 1} \right) \hfill \\
2\left( {{4^{x - 2}} + 9} \right) = 10\left( {{2^{x - 2}} + 1} \right) \hfill \\
2 \cdot {4^{x - 2}} + 18 = 10 \cdot {2^{x - 2}} + 10 \hfill \\
2 \cdot {\left( {{2^{x - 2}}} \right)^2} + 18 = 10 \cdot {2^{x - 2}} + 10 \hfill \\
смена:{2^{x - 2}} = t \hfill \\
2{t^2} + 18 = 10t + 10 \hfill \\
2{t^2} - 10t + 8 = 0 \hfill \\
{t_{1,2}} = \frac{{5 \pm \sqrt {25 - 16} }}{2} \hfill \\
{t_{1,2}} = \frac{{5 \pm 3}}{2} \hfill \\
{t_1} = 1,{t_2} = 4 \hfill \\
{2^{x - 2}} = {1,2^{x - 2}} = 4 \hfill \\
{2^{x - 2}} = 1 \hfill \\
{2^{x - 2}} = {2^0} \hfill \\
x - 2 = 0 \hfill \\
x = 2 \hfill \\
{2^{x - 2}} = 4 \hfill \\
{2^{x - 2}} = {2^2} \hfill \\
x - 2 = 2 \hfill \\
x = 4 \hfill \\
x = \left\{ {2;4} \right\} \hfill \\
\end{gathered} \]

Пр.10

\[\begin{gathered}
\log _{\frac{1}{2}}^24x + {\log _2}\left( {\frac{{{x^2}}}{8}} \right) = 8 \hfill \\
\log _{{2^{ - 1}}}^24x + {\log _2}{x^2} - {\log _2}8 = 8 \hfill \\
{\left( { - {{\log }_2}4x} \right)^2} + 2{\log _2}x - {\log _2}{2^3} = 8 \hfill \\
{\left( {{{\log }_2}4 + {{\log }_2}x} \right)^2} + 2{\log _2}x - 3{\log _2}2 = 8 \hfill \\
{\left( {{{\log }_2}{2^2} + {{\log }_2}x} \right)^2} + 2{\log _2}x - 3{\log _2}2 = 8 \hfill \\
смена:{\log _2}x = t \hfill \\
{\left( {2{{\log }_2}2 + t} \right)^2} + 2t - 3 = 8 \hfill \\
{\left( {2 + t} \right)^2} + 2t - 11 = 0 \hfill \\
4 + 4t + {t^2} + 2t - 11 = 0 \hfill \\
{t^2} + 6t - 7 = 0 \hfill \\
{t_{1,2}} = \frac{{ - 6 \pm \sqrt {36 + 28} }}{2} \hfill \\
{t_{1,2}} = \frac{{ - 6 \pm 8}}{2} \hfill \\
{t_1} = - 7,{t_2} = 1 \hfill \\
{\log _2}x = - 7,{\log _2}x = 1 \hfill \\
\hfill \\
{\log _2}x = - 7 \hfill \\
x = {2^{ - 7}} = \frac{1}{{{2^7}}} \hfill \\
\hfill \\
{\log _2}x = 1 \hfill \\
x = 2 \hfill \\
x = \left\{ {\frac{1}{{{2^7}}};2} \right\} \hfill \\
\end{gathered} \]