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satellite73Best ResponseYou've already chosen the best response.0
domain: you can only take the log of a positive number, so start with \(x+2>0\) and solve for \(x) in one step
 one year ago

satellite73Best ResponseYou've already chosen the best response.0
for the inverse, set \[x=\ln(y+2)3\] and solve for \(y\)
 one year ago

satellite73Best ResponseYou've already chosen the best response.0
this takes three steps \[x=\ln(y+2)3\] \[x+3=\ln(y+2)\] \[e^{x+3}=y+2\] \[y=e^{x+3}2\]
 one year ago

jaersynBest ResponseYou've already chosen the best response.0
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 nonfans apparently
 one year ago

liliyBest ResponseYou've already chosen the best response.0
so is the domain z+, or R ?
 one year ago

liliyBest ResponseYou've already chosen the best response.0
but using an ln graph, anything less than ln(1) is undefined....
 one year ago
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