Question

what would be the inverse function of ln(2+lnx) and it's domain? :)

Answer 1

so do you know how to start finding the inverse function?

Answer 2

yea i guess

Answer 3

so how do you do it?

Answer 4

we interchange x and y and solve the equation for y

Answer 5

good and if you do that to your equation?

Answer 6

that's the problem. I don't under stand how to actually do it for this function

Answer 7

ahhh ok! well we start off with something like this correct:
y=ln(2=lnx)
as the original function

Answer 8

*y = ln(2+lnx)

Answer 9

yea but what after that? sorry but I'm kinda in a hurry :p

Answer 10

I see well after switching we get x = ln(2+lny)
and the part you probably are stuck on it how to switch over the ln (it's just the log scale for e) so we raise both sides by e.
\[e^{x} = 2+\ln y\]
think you can take it from here?

Answer 11

what do you mean by raising both sides by e? raise it to the power e or take natural log on both sides?

Answer 12

oh I'm sorry I meant that we take both sides and make it the power to e. so...
\[e^{x} = e^{\ln (2 + \ln y)}\]

Answer 13

and since lne =1 we get e^x=2+lny ?

Answer 14

yes exactly

Answer 15

sooo now we get y= e^(e^x-2) ? am I right? would this be the inverse?

Answer 16

yup that's correct

Answer 17

now to get the domain and range of inverse functions you simply reverse the two from the original function.

Answer 18

domain of inverse would be range of original function so what would be the range of the original function?

Answer 19

what is the range of just this function y= lnx ?

Answer 20

0 to infinity. I think

Answer 21

that's it's domain, the range is simply all real numbers (think about it's own inverse which is e^x: it's domain is all reals and it's range is [0,infinity)). So since lnx can be anything(aka all real numbers) 2+lnx can be all real numbers as well.

Answer 22

oh sorry I mean (0,infinity) not [0,infinity)

Answer 23

oh yes yes. sorry. So the the domain of our inverse function is simply the set of all real numbers?

Answer 24

well that is the range of the original function correct so it must be and we can check it by looking at the inverse function itself, there are no sqrts or anything of the sort in the function and x is not in the denominator so it makes sense take it's the set of all real numbers.

Answer 25

*that

Answer 26

Oh yes i get it. Thank you SO VERY much! You've been great help.

Answer 27

No problem so do you think you can solve for the range of the inverse or would you like my help?

Answer 28

I think I can do it. And I just needed the domain anyway. But I'll do it just for the sake of it. Thanks again

Answer 29

no problem good luck.

Answer 30

:)