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Lösung Basiswechel

Version 1.1 von Niklas Wunder am 2024/12/18 14:25

a) (\frac{1}{4})^x=(2^{-x})^x=2^{-2x}
b) (\frac{3}{18})^x=(\frac{1}{6})^=6^{-x}
c) 5^{2x+1}=5\cdot (5^2)^x=5\cdot 25^x
d) (\frac{16}{54})^{2x}=(\frac{8}{27})^{2x}=(\frac{2}{3})^{3\cdot 2x}=(-\frac{2}{3})^{-6x}