Respuesta :

konrad509
[tex]f(x)=\log_{\tfrac{1}{4}}x^5\\ y=\log_{\tfrac{1}{4}}x^5\\ \left(\dfrac{1}{4}\right)^y=x^5\\ x=\left(\left(\dfrac{1}{4}\right)^y\right)^{\dfrac{1}{5}}\\ x=\left(\dfrac{1}{4}\right)^{\dfrac{y}{5}}\\x=\left(2^{-2}\right)^{\dfrac{y}{5}}\\x=2^{-\tfrac{2y}{5}}\\ f^{-1}(x)=2^{-\tfrac{2x}{5}}[/tex]


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