Експоненцијалне једначине 2

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Решити експоненцијалну једначину.

пр.5)   $10 \cdot {2^x} - {4^x} = 16$

пр.6)   ${5^{2x - 1}} + {5^{x + 1}} = 250$

пр.7)   ${4^{\frac{2}{x}}} - 5 \cdot {4^{\frac{1}{x}}} =  - 4$

пр.8)   ${4^{\sqrt {x - 2} }} + 16 = 10 \cdot {2^{\sqrt {x - 2} }}$

Пр.5

\[\begin{gathered}
10 \cdot {2^x} - {4^x} = 16 \hfill \\
10 \cdot {2^x} - {\left( {{2^2}} \right)^x} = 16 \hfill \\
10 \cdot {2^x} - {\left( {{2^x}} \right)^2} = 16 \hfill \\
смена:{2^x} = t \hfill \\
10t - {t^2} = 16 \hfill \\
- {t^2} + 10t - 16 = 0 \hfill \\
{t_{1,2}} = \frac{{ - 10 \pm \sqrt {100 - 64} }}{{ - 2}} \hfill \\
{t_{1,2}} = \frac{{ - 10 \pm 6}}{{ - 2}} \hfill \\
{t_1} = 8 \hfill \\
{2^x} = 8 \hfill \\
{2^x} = {2^3} \hfill \\
x = 3 \hfill \\
{t_2} = 2 \hfill \\
{2^x} = 2 \hfill \\
{2^x} = {2^1} \hfill \\
x = 1 \hfill \\
x = \left\{ {1;3} \right\} \hfill \\
\end{gathered} \]

Пр.6

\[\begin{gathered}
{5^{2x - 1}} + {5^{x + 1}} = 250 \hfill \\
{5^{2x}} \cdot {5^{ - 1}} + {5^x} \cdot {5^1} = 250 \hfill \\
\frac{{{5^{2x}}}}{5} + {5^x} \cdot {5^1} = 250 \hfill \\
смена:{5^x} = t \hfill \\
\frac{{{t^2}}}{5} + 5t = 250\left| { \cdot 5} \right. \hfill \\
{t^2} + 25t = 1250 \hfill \\
{t^2} + 25t - 1250 = 0 \hfill \\
{t_{1,2}} = \frac{{ - 25 \pm \sqrt { - 25 + 4 \cdot 1250} }}{2} \hfill \\
{t_{1,2}} = \frac{{ - 25 \pm 75}}{2} \hfill \\
{t_1} = - 50 \hfill \\
{5^x} = - 50 \hfill \\
\end{gathered} \]

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

\[\begin{gathered}
{t_2} = 25 \hfill \\
{5^x} = {5^2} \hfill \\
x = 2 \hfill \\
\end{gathered} \]

Пр.7

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

нема решења

\[\begin{gathered}
{t_2} = 4 \hfill \\
{4^{\frac{1}{x}}} = {4^1} \hfill \\
\frac{1}{x} = 1 \hfill \\
x = 1 \hfill \\
\end{gathered} \]

Пр.8

\[\begin{gathered}
{4^{\sqrt {x - 2} }} + 16 = 10 \cdot {2^{\sqrt {x - 2} }} \hfill \\
{\left( {{2^2}} \right)^{\sqrt {x - 2} }} + 16 = 10 \cdot {2^{\sqrt {x - 2} }} \hfill \\
смена:{2^{\sqrt {x - 2} }} = t \hfill \\
{t^2} + 16 = 10t \hfill \\
{t^2} - 10t + 16 = 0 \hfill \\
{t_{1,2}} = \frac{{10 \pm \sqrt {100 - 64} }}{2} \hfill \\
{t_{1,2}} = \frac{{10 \pm 6}}{2} \hfill \\
{t_1} = 2 \hfill \\
{2^{\sqrt {x - 2} }} = {2^1} \hfill \\
\sqrt {x - 2} = 1 \Leftrightarrow \left\{ \begin{gathered}
x - 2 \geqslant 0 \hfill \\
x - 2 = 1 \hfill \\
\end{gathered} \right. \Leftrightarrow \left\{ \begin{gathered}
x \geqslant 2 \hfill \\
x = 3 \hfill \\
\end{gathered} \right. \hfill \\
{t_2} = 8 \hfill \\
{2^{\sqrt {x - 2} }} = {2^3} \hfill \\
\sqrt {x - 2} = 3 \Leftrightarrow \left\{ \begin{gathered}
x - 2 \geqslant 0 \hfill \\
x - 2 = 9 \hfill \\
\end{gathered} \right. \Leftrightarrow \left\{ \begin{gathered}
x \geqslant 2 \hfill \\
x = 11 \hfill \\
\end{gathered} \right. \hfill \\
x = \left\{ {3;11} \right\} \hfill \\
\end{gathered} \]