Dada a função f(x)= 2+log_0,25 (x+2). Qual a função inversa dela?
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Olá, Lucas.
![f(x)= 2+\log_{0,25}(x+2)\Rightarrow x= 2+\log_{4^{-1}}[f^{-1}(x)+2]\Rightarrow \\\\
x-2=\frac{\log[f^{-1}(x)+2]}{\log{4^{-1}}}\Rightarrow x-2=\frac{\log[f^{-1}(x)+2]}{-\log4}\Rightarrow\\\\
(-\log4)(x-2)=\log[f^{-1}(x)+2]}\Rightarrow f^{-1}(x)+2 = 10^{(\log4)(2-x)}\Rightarrow\\\\
\boxed{f^{-1}(x)=4^{2-x}-2}](https://tex.z-dn.net/?f=f%28x%29%3D+2%2B%5Clog_%7B0%2C25%7D%28x%2B2%29%5CRightarrow+x%3D+2%2B%5Clog_%7B4%5E%7B-1%7D%7D%5Bf%5E%7B-1%7D%28x%29%2B2%5D%5CRightarrow+%5C%5C%5C%5C%0Ax-2%3D%5Cfrac%7B%5Clog%5Bf%5E%7B-1%7D%28x%29%2B2%5D%7D%7B%5Clog%7B4%5E%7B-1%7D%7D%7D%5CRightarrow+x-2%3D%5Cfrac%7B%5Clog%5Bf%5E%7B-1%7D%28x%29%2B2%5D%7D%7B-%5Clog4%7D%5CRightarrow%5C%5C%5C%5C%0A%28-%5Clog4%29%28x-2%29%3D%5Clog%5Bf%5E%7B-1%7D%28x%29%2B2%5D%7D%5CRightarrow+f%5E%7B-1%7D%28x%29%2B2+%3D+10%5E%7B%28%5Clog4%29%282-x%29%7D%5CRightarrow%5C%5C%5C%5C%0A%5Cboxed%7Bf%5E%7B-1%7D%28x%29%3D4%5E%7B2-x%7D-2%7D "f(x)= 2+\log_{0,25}(x+2)\Rightarrow x= 2+\log_{4^{-1}}[f^{-1}(x)+2]\Rightarrow \\\\
x-2=\frac{\log[f^{-1}(x)+2]}{\log{4^{-1}}}\Rightarrow x-2=\frac{\log[f^{-1}(x)+2]}{-\log4}\Rightarrow\\\\
(-\log4)(x-2)=\log[f^{-1}(x)+2]}\Rightarrow f^{-1}(x)+2 = 10^{(\log4)(2-x)}\Rightarrow\\\\
\boxed{f^{-1}(x)=4^{2-x}-2}")
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