Opgave 1
$
\begin{array}{l}
0,25^{2t - 1} = 256\sqrt 2 \\
\left( {\frac{1}{4}} \right)^{2t - 1} = 2^8 \cdot 2^{\frac{1}{2}} \\
\left( {\frac{1}{{2^2 }}} \right)^{2t - 1} = 2^{8\frac{1}{2}} \\
\left( {2^{ - 2} } \right)^{2t - 1} = 2^{8\frac{1}{2}} \\
2^{ - 4t + 2} = 2^{8\frac{1}{2}} \\
- 4t + 2 = 8\frac{1}{2} \\
- 4t = 6\frac{1}{2} \\
t = - \frac{{13}}{8} \\
t = - 1\frac{5}{8} \\
\end{array}
$
Opgave 2
$
\begin{array}{l}
2\left\{ {{}^2\log (x - 1) - {}^4\log (x^2 + 1)} \right\} = 1 - {}^4\log (25) \\
2\left\{ {{}^2\log (x - 1) - \frac{1}{2} \cdot {}^2\log (x^2 + 1)} \right\} = {}^2\log (2) - {}^2\log (5) \\
2 \cdot {}^2\log (x - 1) - {}^2\log (x^2 + 1) = {}^2\log (\frac{2}{5}) \\
{}^2\log (\left( {x - 1} \right)^2 ) - {}^2\log (x^2 + 1) = {}^2\log (\frac{2}{5}) \\
{}^2\log (\frac{{\left( {x - 1} \right)^2 }}{{x^2 + 1}}) = {}^2\log (\frac{2}{5}) \\
\frac{{\left( {x - 1} \right)^2 }}{{x^2 + 1}} = \frac{2}{5} \\
5\left( {x - 1} \right)^2 = 2\left( {x^2 + 1} \right) \\
5\left( {x^2 - 2x + 1} \right) = 2x^2 + 2 \\
5x^2 - 10x + 5 = 2x^2 + 2 \\
3x^2 - 10x + 3 = 0 \\
\left( {3x - 1} \right)\left( {x - 3} \right) = 0 \\
3x - 1 = 0 \vee x - 3 = 0 \\
x = \frac{1}{3}\left( {v.n.} \right) \vee x = 3 \\
x = 3 \\
\end{array}
$
Opgave 3
$
\begin{array}{l}
\left( {\frac{1}{9}} \right)^{x + 4} = \frac{2}{{\sqrt 3 }} \\
\left( {\left( {\frac{1}{3}} \right)^2 } \right)^{x + 4} = 2 \cdot 3^{^{ - \frac{1}{2}} } \\
\left( {\left( {3^{ - 1} } \right)} \right)^{2x + 8} = 2 \cdot 3^{^{ - \frac{1}{2}} } \\
3^{ - 2x - 8} = 2 \cdot 3^{^{ - \frac{1}{2}} } \\
{}^3\log \left( {3^{ - 2x - 8} } \right) = {}^3\log \left( {2 \cdot 3^{^{ - \frac{1}{2}} } } \right) \\
\log \left( {3^{ - 2x - 8} } \right) = {}^3\log \left( 2 \right) + {}^3\log \left( {3^{^{ - \frac{1}{2}} } } \right) \\
- 2x - 8 = {}^3\log \left( 2 \right) - \frac{1}{2} \\
- 2x = {}^3\log \left( 2 \right) + 7\frac{1}{2} \\
x = - \frac{1}{2} \cdot {}^3\log \left( 2 \right) - 3\frac{3}{4} \\
\end{array}
$
