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Given the function f(x) = 0.3(4)^x, what is the value of f^−1(6)? a: 0.360 b:1.242 c:2.161 d:2.312
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to find the inverse "relation" you'd simply swap about the variables, so \(\bf f(x) ={\color{red}{ y}} =0.3(4)^{\color{blue}{ x}}\qquad inverse\implies {\color{blue}{ x}}=0.3(4)^{\color{red}{ y}} \\ \quad \\ \textit{log cancellation rule of }{\color{blue}{ log_aa^x=x}}\qquad thus \\ \quad \\ {\color{blue}{ x}}=0.3(4)^{\color{red}{ y}}\implies \cfrac{x}{0.3}=4^y\implies log_4\left(\cfrac{x}{0.3}\right)=log_4(4^y) \\ \quad \\ log_4\left(\cfrac{x}{0.3}\right)=y \iff f^{-1} \\ \quad \\ f^{-1}(16)=log_4\left(\cfrac{16}{0.3}\right)\)
hmm 6
\(\bf f^{-1}(6)=log_4\left(\cfrac{6}{0.3}\right)\)
So, would that be a?
@jdoe0001
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