Lösung Basiswechel

\(f(x)=(\frac{1}{4})^x\), neue Basis \(b=2\)
\(\frac{1}{4} = 2^{-2} \Rightarrow f(x)=2^{-2x}\)

\(f(x)=9^x\), neue Basis \(b=\frac{1}{3}\)
\(9 = (\frac{1}{3})^{-2} \Rightarrow f(x)=(\frac{1}{3})^{-2x}\)

\(f(x)=5^{2x+1}\), neue Basis \(b=25\)
\(5 = 25^{\frac{1}{2}} \Rightarrow f(x)=25^{\frac{1}{2}\cdot(2x+1)}=25^{x+\frac{1}{2}}\)