Here's the question you clicked on:
liliy
find the domain and the inverse of the function: f(x)=ln(x+2)-3
domain: you can only take the log of a positive number, so start with \(x+2>0\) and solve for \(x) in one step
for the inverse, set \[x=\ln(y+2)-3\] and solve for \(y\)
this takes three steps \[x=\ln(y+2)-3\] \[x+3=\ln(y+2)\] \[e^{x+3}=y+2\] \[y=e^{x+3}-2\]
Hey satellite, just out of curiosity, what do you do for a living? Sorry to barge in here, but you don't accept messages from non-fans apparently
so is the domain z+, or R ?
but using an ln graph, anything less than ln(1) is undefined....