Текст задатака објашњених у видео лекцији.

Решити експоненцијалну једначину.

пр.1)   ${27^x} = \frac{1}{9}$

пр.2)   ${a^x} \cdot \sqrt a  = {a^{\frac{3}{x}}}$    $a > 0$, $a \ne 1$

пр.3)   $0,125 \cdot {4^{2x - 3}} = {\left( {\frac{8}{{\sqrt 2 }}} \right)^x}$

пр.4)   $0,{5^{ - 3}} \cdot 2 \cdot {8^{2x - 3}} = {\left( {\frac{{\sqrt[3]{2}}}{4}} \right)^{ - 2x}}$

Пр.1

\[\begin{gathered}
{27^x} = \frac{1}{9} \hfill \\
{\left( {{3^3}} \right)^x} = \frac{1}{{{3^2}}} \hfill \\
{3^{3x}} = {3^{ - 2}} \hfill \\
3x = - 2 \hfill \\
x = - \frac{2}{3} \hfill \\
\end{gathered} \]

Пр.2

\[\begin{gathered}
{a^x} \cdot \sqrt a = {a^{\frac{3}{x}}} \hfill \\
a > 0 \hfill \\
a \ne 1 \hfill \\
{a^x} \cdot {a^{\frac{1}{2}}} = {a^{^{\frac{3}{x}}}} \hfill \\
{a^{x + \frac{1}{2}}} = {a^{^{\frac{3}{x}}}} \hfill \\
x + \frac{1}{2} = \frac{3}{x}\left| { \cdot 2x} \right. \hfill \\
2{x^2} + x = 6 \hfill \\
2{x^2} + x - 6 = 0 \hfill \\
{x_{1,2}} = \frac{{ - 1 \pm \sqrt {49} }}{4} \hfill \\
{x_{1,2}} = \frac{{ - 1 \pm 7}}{4} \hfill \\
{x_1} = - 2,{x_2} = \frac{3}{2} \hfill \\
\end{gathered} \]

Пр.3

грешка видео

\[\begin{gathered}
0,125 \cdot {4^{2x - 3}} = {\left( {\frac{8}{{\sqrt 2 }}} \right)^x} \hfill \\
0,125 = \frac{{125}}{{1000}} = \frac{1}{8} \hfill \\
\frac{1}{8} \cdot {4^{2x - 3}} = {\left( {\frac{8}{{\sqrt 2 }}} \right)^x} \hfill \\
\frac{1}{{{2^3}}} \cdot {4^{2x - 3}} = {\left( {\frac{8}{{\sqrt 2 }}} \right)^x} \hfill \\
{2^{ - 3}} \cdot {\left( {{2^2}} \right)^{2x - 3}} = {\left( {\frac{{{2^3}}}{{{2^{\frac{1}{2}}}}}} \right)^x} \hfill \\
{2^{ - 3}} \cdot {2^{4x - 6}} = {\left( {{2^{3 - \frac{1}{2}}}} \right)^x} \hfill \\
{2^{ - 3 + 4x - 6}} = {\left( {{2^{^{\frac{5}{2}}}}} \right)^x} \hfill \\
{2^{4x - 9}} = {2^{\frac{5}{2}x}} \hfill \\
4x - 9 = \frac{5}{2}x\left| { \cdot 2} \right. \hfill \\
8x - 18 = 5x \hfill \\
3x = 18 \hfill \\
x = 6 \hfill \\
\end{gathered} \]

Пр.4

\[\begin{gathered}
{0,5^{ - 3}} \cdot 2 \cdot {8^{2x - 3}} = {\left( {\frac{{\sqrt[3]{2}}}{4}} \right)^{ - 2x}} \hfill \\
{\left( {\frac{1}{2}} \right)^{ - 3}} \cdot {2^1} \cdot {\left( {{2^3}} \right)^{2x - 3}} = {\left( {\frac{{{2^{\frac{1}{3}}}}}{{{2^2}}}} \right)^{ - 2x}} \hfill \\
{2^3} \cdot {2^1} \cdot {2^{6x - 9}} = {\left( {{2^{\frac{1}{3} - 2}}} \right)^{ - 2x}} \hfill \\
{2^{3 + 1 + 6x - 9}} = {\left( {{2^{ - \frac{5}{3}}}} \right)^{ - 2x}} \hfill \\
{2^{6x - 5}} = {2^{\frac{{10}}{3}x}} \hfill \\
6x - 5 = \frac{{10}}{3}x\left| { \cdot 3} \right. \hfill \\
18x - 15 = 10x \hfill \\
8x = 15 \hfill \\
x = \frac{{15}}{8} \hfill \\
\end{gathered} \]