Call Now Button
Други разред средње школе

Логаритамске једначине 1

Задаци

Текст задатака објашњених у видео лекцији.

Решити логаритамску једначину.

пр.1)   ${\log _{\frac{1}{3}}}\left( {x - 1} \right) = 2$

пр.2)   ${\log _{x - 1}}3 = 2$

пр.3)   ${\log _2}{\log _4}{\log _3}x =  - 1$

Пр.1

\[\begin{gathered}
{\log _{\frac{1}{3}}}\left( {x - 1} \right) = 2 \hfill \\
\boxed{{{\log }_a}b = x,a > 0,a \ne 1,b > 0 \Leftrightarrow {a^x} = b} \hfill \\
{\log _{\frac{1}{3}}}\left( {x - 1} \right) = 2,x - 1 > 0 \Rightarrow x > 1 \hfill \\
{\left( {\frac{1}{3}} \right)^2} = x - 1 \hfill \\
\frac{1}{9} = x - 1 \hfill \\
x = \frac{1}{9} + 1 \hfill \\
x = \frac{{10}}{9} \hfill \\
\end{gathered} \]

Пр.2

 

\[\begin{gathered}
{\log _{x - 1}}3 = 2 \hfill \\
x - 1 > 0,x - 1 \ne 1 \hfill \\
x > 1,x \ne 2 \hfill \\
{\left( {x - 1} \right)^2} = 3 \hfill \\
{x^2} - 2x + 1 - 3 = 0 \hfill \\
{x^2} - 2x - 2 = 0 \hfill \\
{x_{1,2}} = \frac{{2 \pm \sqrt {4 + 8} }}{2} \hfill \\
{x_{1,2}} = \frac{{2 \pm 2\sqrt 3 }}{2} \hfill \\
{x_1} = 1 - \sqrt 3 ,{x_2} = 1 + \sqrt 3 \hfill \\
\end{gathered} \]

Знамо да je $x > 1$ и $x \ne 2.$ Онда је решење jдначине  ${x_2} = 1 + \sqrt 3$

Пр.3

\[\begin{gathered}
{\log _2}{\log _4}{\log _3}x = - 1 \hfill \\
\boxed{{{\log }_a}b = x,a > 0,a \ne 1,b > 0 \Leftrightarrow {a^x} = b} \hfill \\
x > 0,{\log _3}x > 0,{\log _4}{\log _3}x > 0 \hfill \\
\hfill \\
{2^{ - 1}} = {\log _4}{\log _3}x \hfill \\
{\log _4}{\log _3}x = \frac{1}{2} \hfill \\
{4^{\frac{1}{2}}} = {\log _3}x \hfill \\
{\log _3}x = 2 \hfill \\
{3^2} = x \hfill \\
x = 9 \hfill \\
\end{gathered} \]

Call Now Button