Question

Use the one to one correspondences
ln: (0, infinity) --> R
f: ( 2, infinity) --> (0, infinity)
Where f(x)=x-2
to describe a one to one correspondence from (0,infinity) onto (2, infinity)

Answer 1

Umm I so dont follow what to do here

Answer 2

on first glace it looks like "first take the log, then subtract 2"

Answer 3

hmm maybe "add 2"?

Answer 4

idk

Answer 5

like i am a lil stimped

Answer 6

stumped*

Answer 7

yeah this is weird
if you want a one to one corresponedence between \((2,\infty)\) and \((0,\infty)\) what is wrong with \(x:\to x-2\)

Answer 8

Nothing -_-

Answer 9

I mean is that all that needs to be done? This is nerdy

Answer 10

Umm how about from (2, infinity) onto R
x: ln(x-2)?

Answer 11

that would certainly do it

Answer 12

yayyyy, Its weird that it is soo simple but i guess i wont complain

Answer 13

sometimes it is so simple it seems hard

Answer 14

Now from R onto (2, infinity)
wld it be e^(x-2)?

Answer 15

@nbouscal is this correct?

Answer 16

@mukushla

Answer 17

lol....idk...

Answer 18

thanks :)

Answer 19

@satellite73 can u check this?

Answer 20

Here's what I would do. Find a 1-1 correspondence (a bijection) from \((2,\infty)\) to \((0,\infty)\). Then take the inverse.

Answer 21

umm so i did the other examples incorrect too?

Answer 22

Let me read over this more carefully.

Answer 23

So satellite went from (2, infinity) to (0, infinity) at first. We need to go the other way. It's not much harder. You just need to use \(f^{-1}=x+2\).

Answer 24

From (2, infinity) to R, you nailed it. It's just \(\ln(f(x))=\ln(x-2)\).

Answer 25

From R to (2, infinity), it's the inverse of what I just wrote above. \[f^{-1}(\ln^{-1}(x))=f^{-1}(e^x)=e^x+2\]

Answer 26

Alrrighhtttttt. YAYYYYYYY

Answer 27

At

Answer 28

least i accomplished smth this evening

Answer 29

Thanks :)

Answer 30

You're welcome.