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for the inverse, set
\[x=\ln(y+2)-3\] and solve for \(y\)

this takes three steps
\[x=\ln(y+2)-3\]
\[x+3=\ln(y+2)\]
\[e^{x+3}=y+2\]
\[y=e^{x+3}-2\]

so is the domain z+, or R ?

neither

but using an ln graph, anything less than ln(1) is undefined....