Sideversjonen fra 29. des. 2018 kl. 15:37

DEL 1

$x²-3x+1=3x+8 \\ x²-6x-7=0 \\ x=\frac{6\pm\sqrt{(-6)²-4\cdot(-7)}}{2} \\ x=\frac{6\pm 8}{2} \\ x_1=-1 \vee x_2=7$

$lg(x⁴)-lg(x³)+lg(x²)-lg\,x=6 \\ 4\,lg\,x-3\,lg\,x+2\,lg\,x-lg\,x=6 \\ 2\,lg\,x=6 \\ lg\,x=3 \\ x=10³ \\ x=1000$

$10\cdot 4^x=5\cdot 2^x \\ \frac{2^{2x}}{2^x} = \frac{5}{10} \\ 2^{2x-x} = \frac{1}{2} \\ 2^x = 2^(-1) \\ x=-1$

@@ Linje 18: / Linje 18: @@
 ===c)===
+$10\cdot 4^x=5\cdot 2^x \\ \frac{2^{2x}}{2^x} = \frac{5}{10} \\ 2^{2x-x} = \frac{1}{2} \\ 2^x = 2^(-1) \\ x=-1$
+==Oppgave 2==
+===a)===