e)
[tex]f:\mathbb{R}\to (0,\infty),\quad f(x) = \log_{2}(3^x+1)\\ \\ \boxed{f^{-1}\Big(f(x)\Big) = x} \to \text{Teorema fundamentala a functiei inverse.} \\\\\\ f^{-1}\Big(f(y)\Big) = y\\\\ f(y) = x \Rightarrow \log_{2}(3^y+1) = x \Rightarrow 3^y+1 = 2^x \Rightarrow 3^y = 2^x-1 \Rightarrow \\ \\ \Rightarrow y = \log_{3} {(2^x-1)} \\ \\ \Rightarrow f^{-1}(x) = \log_{3}(2^x-1)\\ \\ \Rightarrow \boxed{f:(0,\infty)\to \mathbb{R},\quad f^{-}(x) = \log_{3}(2^x-1)}[/tex]
f)
[tex]f:(1,\infty)\to (0,\infty),\quad f(x) = 3\log_{2}x \\ \\ f^{-1}\Big(f(y)\Big) = y \\ \\ f(y) = x \Rightarrow 3\log_{2}y = x \Rightarrow \log_{2}y = \dfrac{x}{3} \Rightarrow y = 2^{\dfrac{x}{3}} \Rightarrow \\ \\ \Rightarrow y = \sqrt[3]{2^x} \Rightarrow f^{-1}(x) = \sqrt[3]{2^x}\\ \\ \Rightarrow \boxed{f:(0,\infty)\to (1,\infty),\quad f^{-1}(x) = \sqrt[3]{2^x}}[/tex]
