Question

Given the function f(x) = 0.5(3)x, what is the value of f−1(7)?

Answer 1

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Explanation:

Answer 2

`\bf \textit{we do the same here, the \underline{switcharoo} of the variables} \\\\\\ f(x)=y=0.5(3)^x\qquad inverse\implies \boxed{x}=0.5(3)^{\boxed{y}}\\\\ -------------------------------\\\\ \textit{Logarithm of exponentials}\\\\ log_{{ a}}\left( x^{{ b}} \right)\implies {{ b}}\cdot log_{{ a}}(x)\impliedby \textit{we'll use this}\\\\ -------------------------------\\\\`


`\bf x=0.5(3)^y\implies \cfrac{x}{0.5}=3^y\implies log\left( (x)/(0.5) \right)=log(3^y) \\\\\\ log\left( (x)/(0.5) \right)=y\cdot log(3)\implies \cfrac{log\left( (x)/(0.5) \right)}{log(3)}=y\impliedby f^(-1) \\\\\\ thus\qquad \cfrac{log\left( (7)/(0.5) \right)}{log(3)}=f^(-1)(7)`

and surely you know how much that'd be.