Lösung Basiswechsel
Version 1.1 von Martin Rathgeb am 2025/04/23 23:12
\(f(x)=(\frac{1}{4})^x\), 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 \(b=25\)
\[5 = 25^{\frac{1}{2}} \Rightarrow f(x)=25^{\frac{1}{2}\cdot(2x+1)}=25^{x+\frac{1}{2}}\]