<s>[INST] How do I find the inverse of this exponential function? $x=-3(3^{-x})+9$
I know the steps up until a certain point.
$x=-3(3^{-y})+9$
$x-9=-3(3^{-y})$
$\frac{(x-9)}{-3} = 3^y$
$ln (\frac{x-9}{-3}) = -y * ln 3$
Not sure what to do from here. I know I have to get y by itself but thats it. Can anyone help please?
 [/INST] We want to solve for $y$, after switching $x$ and $y$.
$y=-3(3^{-x})+9$ 
becomes 
$x=-3(3^{-y})+9\\
\implies x-9=-3(3^{-y})\\
\implies \dfrac{x-9}{-3}=3^{-y}\\
\implies -\dfrac{1}{3}x+3=3^{-y}\\
\implies \ln\left(-\dfrac{1}{3}x+3\right)=\ln (3^{-y})\\
\implies \ln\left(-\dfrac{1}{3}x+3\right)=-y\cdot \ln 3\\
\implies -\dfrac{ \ln\left(-\dfrac{1}{3}x+3\right)}{\ln 3}=y$
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