[tex]\bf \textit{we do the same here, the \underline{switcharoo} of the variables}
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f(x)=y=0.5(3)^x\qquad inverse\implies \boxed{x}=0.5(3)^{\boxed{y}}\\\\
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\textit{Logarithm of exponentials}\\\\
log_{{ a}}\left( x^{{ b}} \right)\implies {{ b}}\cdot log_{{ a}}(x)\impliedby \textit{we'll use this}\\\\
-------------------------------\\\\[/tex]
[tex]\bf x=0.5(3)^y\implies \cfrac{x}{0.5}=3^y\implies log\left( \frac{x}{0.5} \right)=log(3^y)
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log\left( \frac{x}{0.5} \right)=y\cdot log(3)\implies \cfrac{log\left( \frac{x}{0.5} \right)}{log(3)}=y\impliedby f^{-1}
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thus\qquad \cfrac{log\left( \frac{7}{0.5} \right)}{log(3)}=f^{-1}(7)[/tex]
and surely you know how much that'd be.
