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Логаритамске једначине 2

Задаци

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Решити логаритамску једначину.

пр.4)   ${\log _2} {\log _2}x = {\log _2}3 + {\log _2}4$

пр.5)   $2{\left( {{{\log }_x}\sqrt 5 } \right)^2} - 3{\log _x}\sqrt 5  + 1 = 0$

пр.6)   ${\log _x}2 - {\log _4}x + \frac{7}{6} = 0$

Пр.4

\[\begin{gathered}
{\log _2}{\log _2}x = {\log _2}3 + {\log _2}4 \hfill \\
x > 0,{\log _2}x > 0 \hfill \\
{\log _2}{\log _2}x = {\log _2}3 \cdot 4 \hfill \\
{\log _2}x = 12 \hfill \\
x = {2^{12}} \hfill \\
\end{gathered} \]

Пр.5

\[\begin{gathered}
2{\left( {{{\log }_x}\sqrt 5 } \right)^2} - 3{\log _x}\sqrt 5 + 1 = 0 \hfill \\
x > 0,x \ne 1 \hfill \\
смена:{\log _x}\sqrt 5 = t \hfill \\
2{t^2} - 3t + 1 = 0 \hfill \\
{t_{1,2}} = \frac{{3 \pm \sqrt {9 - 8} }}{4} \hfill \\
{t_{1,2}} = \frac{{3 \pm 1}}{4} \hfill \\
{t_1} = \frac{1}{2},{t_2} = 1 \hfill \\
\hfill \\
{\log _x}\sqrt 5 = \frac{1}{2} \hfill \\
{x^{\frac{1}{2}}} = \sqrt 5 \hfill \\
x = 5 \hfill \\
{\log _x}\sqrt 5 = 1 \hfill \\
{x^1} = \sqrt 5 \hfill \\
x = \sqrt 5 \hfill \\
\end{gathered} \]

Пр.6

\[\begin{gathered}
{\log _x}2 - {\log _4}x + \frac{7}{6} = 0 \hfill \\
x > 0,x \ne 1 \hfill \\
\frac{1}{{{{\log }_2}x}} - {\log _{{2^2}}}x + \frac{7}{6} = 0 \hfill \\
\frac{1}{{{{\log }_2}x}} - \frac{1}{2}{\log _2}x + \frac{7}{6} = 0 \hfill \\
смена:{\log _2}x = t \hfill \\
\frac{1}{t} - \frac{1}{2}t + \frac{7}{6} = 0\left| { \cdot 6} \right.t \hfill \\
- 3{t^2} + 7t + 6 = 0 \hfill \\
{t_{1,2}} = \frac{{ - 7 \pm \sqrt {49 + 72} }}{{ - 6}} \hfill \\
{t_{1,2}} = \frac{{ - 7 \pm 11}}{{ - 6}} \hfill \\
{t_1} = 3,{t_2} = - \frac{2}{3} \hfill \\
\hfill \\
{\log _2}x = 3 \hfill \\
x = {2^3} \hfill \\
x = 8 \hfill \\
{\log _2}x = - \frac{2}{3} \hfill \\
x = {2^{ - \frac{2}{3}x}} \hfill \\
x = \frac{1}{{\sqrt[3]{{{2^2}}}}} = \frac{1}{{\sqrt[3]{4}}} \hfill \\
\end{gathered} \]

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