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}