\require{AMSmath} Uitwerkingen Opgave 1 $\begin{array}{l} a. \\ 2^x = 6 \\ x = {}^2\log (6) \\ \\ b. \\ 7^{2x + 5} = 2^3 \\ 7^{2x + 5} = 8 \\ 2x + 5 = {}^7\log (8) \\ 2x = {}^7\log (8) - 5 \\ x = \frac{1}{2} \cdot {}^7\log (8) - 2\frac{1}{2} \\ \\ c. \\ 11^{x^2 - 4x} = 121 \\ 11^{x^2 - 4x} = 11^2 \\ x^2 - 4x = 2 \\ (x - 2)^2 - 4 = 2 \\ (x - 2)^2 = 6 \\ x - 2 = \pm \sqrt 6 \\ x = 2 \pm \sqrt 6 \\ \end{array}$ ©2004-2024 WisFaq
\require{AMSmath}
Opgave 1 $\begin{array}{l} a. \\ 2^x = 6 \\ x = {}^2\log (6) \\ \\ b. \\ 7^{2x + 5} = 2^3 \\ 7^{2x + 5} = 8 \\ 2x + 5 = {}^7\log (8) \\ 2x = {}^7\log (8) - 5 \\ x = \frac{1}{2} \cdot {}^7\log (8) - 2\frac{1}{2} \\ \\ c. \\ 11^{x^2 - 4x} = 121 \\ 11^{x^2 - 4x} = 11^2 \\ x^2 - 4x = 2 \\ (x - 2)^2 - 4 = 2 \\ (x - 2)^2 = 6 \\ x - 2 = \pm \sqrt 6 \\ x = 2 \pm \sqrt 6 \\ \end{array}$
$\begin{array}{l} a. \\ 2^x = 6 \\ x = {}^2\log (6) \\ \\ b. \\ 7^{2x + 5} = 2^3 \\ 7^{2x + 5} = 8 \\ 2x + 5 = {}^7\log (8) \\ 2x = {}^7\log (8) - 5 \\ x = \frac{1}{2} \cdot {}^7\log (8) - 2\frac{1}{2} \\ \\ c. \\ 11^{x^2 - 4x} = 121 \\ 11^{x^2 - 4x} = 11^2 \\ x^2 - 4x = 2 \\ (x - 2)^2 - 4 = 2 \\ (x - 2)^2 = 6 \\ x - 2 = \pm \sqrt 6 \\ x = 2 \pm \sqrt 6 \\ \end{array}$
©2004-2024 WisFaq