\require{AMSmath}

Uitwerkingen

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}
$


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