Експоненцијалне једначине 3

Текст задатака објашњених у видео лекцији.

Решити експоненцијалну једначину.

пр.9)    ${9^x} + {6^x} = 2 \cdot {4^x}$

пр.10)  $3 \cdot {16^x} + 2 \cdot {81^x} = 5 \cdot {36^x}$

Пр.9 

\[\begin{gathered}
{9^x} + {6^x} = 2 \cdot {4^x} \hfill \\
{\left( {{3^2}} \right)^x} + {\left( {3 \cdot 2} \right)^x} = 2 \cdot {\left( {{2^2}} \right)^x} \hfill \\
{3^{2x}} + {3^x} \cdot {2^x} = 2 \cdot {2^{2x}}\left| \div \right.{2^{2x}} \hfill \\
\frac{{{3^{2x}}}}{{{2^{2x}}}} + \frac{{{3^x} \cdot {2^x}}}{{{2^{2x}}}} = 2 \hfill \\
{\left( {\frac{3}{2}} \right)^{2x}} + {\left( {\frac{3}{2}} \right)^x} = 2 \hfill \\
смена:{\left( {\frac{3}{2}} \right)^x} = t \hfill \\
{t^2} + t - 2 = 0 \hfill \\
{t_{1,2}} = \frac{{ - 1 \pm \sqrt {1 + 8} }}{2} \hfill \\
{t_{1,2}} = \frac{{ - 1 \pm 3}}{2} \hfill \\
{t_1} = - 2 \hfill \\
{\left( {\frac{3}{2}} \right)^x} = - 2 \hfill \\
\end{gathered} \]

ова експоненцијална једначина нема решења (експоненцијална функција је увек позитивна)

\[\begin{gathered}
{t_2} = 1 \hfill \\
{\left( {\frac{3}{2}} \right)^x} = 1 \hfill \\
{\left( {\frac{3}{2}} \right)^x} = {\left( {\frac{3}{2}} \right)^0} \hfill \\
x = 0 \hfill \\
\end{gathered} \]

Пр.10

\[\begin{gathered}
3 \cdot {16^x} + 2 \cdot {81^x} = 5 \cdot {36^x} \hfill \\
3 \cdot {4^{2x}} + 2 \cdot {9^{2x}} = 5{\left( {4 \cdot 9} \right)^x} \hfill \\
3 \cdot {4^{2x}} + 2 \cdot {9^{2x}} = 5 \cdot {4^x} \cdot {9^x}\left| { \div {9^{2x}}} \right. \hfill \\
3 \cdot \frac{{{4^{2x}}}}{{{9^{2x}}}} + 2 \cdot \frac{{{9^{2x}}}}{{{9^{2x}}}} = \frac{{5 \cdot {4^x} \cdot {9^x}}}{{{9^{2x}}}} \hfill \\
3 \cdot {\left( {\frac{4}{9}} \right)^{2x}} + 2 = 5{\left( {\frac{4}{9}} \right)^x} \hfill \\
смена:{\left( {\frac{4}{9}} \right)^x} = t \hfill \\
3{t^2} + 2 = 5t \hfill \\
3{t^2} - 5t + 2 = 0 \hfill \\
{t_{1,2}} = \frac{{5 \pm \sqrt {25 - 24} }}{6} \hfill \\
{t_{1,2}} = \frac{{5 \pm 1}}{6} \hfill \\
{t_1} = 1 \hfill \\
{\left( {\frac{4}{9}} \right)^x} = 1 \hfill \\
{2^x} = {\left( {\frac{4}{9}} \right)^0} \hfill \\
x = 0 \hfill \\
{t_2} = \frac{2}{3} \hfill \\
{\left( {\frac{4}{9}} \right)^x} = \frac{2}{3} \hfill \\
{\left( {\frac{2}{3}} \right)^{2x}} = \frac{2}{3} \hfill \\
2x = 1 \hfill \\
x = \frac{1}{2} \hfill \\
x = \left\{ {0;\frac{1}{2}} \right\} \hfill \\
\end{gathered} \]