1.7 Lösning 6b

$$\left({4^{40} \over 4} \; \Big / \; 4^{38}\right)^{-1} \, = \, \left({4^{40} \over 4\,^1} \; \Big / \; 4^{38}\right)^{-1} \, = \, \left(4^{39} \; / \; 4^{38}\right)^{-1} \, = \,$$

$$\, = \, \left(4^{39-38}\right)^{-1} \, = \, 4\,^{-1} \, = \, {1 \over 4} \, = \, 0,25$$