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# Inverse of exponential functions

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1. I havent really ever been taught how to find the inverse of an exponential function but have been asked to find the inverse of f(x)= 1-2^x

Can anyone help?
2. So if you have so here then the inverse is a reflection in the line y=x, so all you do is swap the x and y variables around so the inverse has equation . Normally though they want the equation in the form so just rearrange to get the equation in x.
3. (Original post by langnerd)
I havent really ever been taught how to find the inverse of an exponential function but have been asked to find the inverse of f(x)= 1-2^x

Can anyone help?
Swap the x's and y's for inverse, rearrange for y. Why do this? Well as B_9710 said, y=x is a line of reflection and inverses are reflections along that line. These must be one-to-one functions though, so you can't do it for something like unless you restrict the domain.
4. (Original post by langnerd)
I havent really ever been taught how to find the inverse of an exponential function but have been asked to find the inverse of f(x)= 1-2^x

Can anyone help?
Log base 2 comes to mind
5. (Original post by B_9710)
So if you have so here then the inverse is a reflection in the line y=x, so all you do is swap the x and y variables around so the inverse has equation . Normally though they want the equation in the form so just rearrange to get the equation in x.
Thank you all for your help. How exactly can I isolate y and make it the subject as the answers I have come up with so far are not a reflection in the line y=x?
6. (Original post by langnerd)
Thank you all for your help. How exactly can I isolate y and make it the subject as the answers I have come up with so far are not a reflection in the line y=x?
Just rearrange to make y the subject.
7. (Original post by langnerd)
Thank you all for your help. How exactly can I isolate y and make it the subject as the answers I have come up with so far are not a reflection in the line y=x?
Get y on one side then use a logarithm with an appropriate base.
8. (Original post by RDKGames)
Get y on one side then use a logarithm with an appropriate base.
I got y=log2(1-x) which isn't a reflection?
9. (Original post by langnerd)
I got y=log2(1-x) which isn't a reflection?
It is in the line y=x.
10. (Original post by langnerd)
I got y=log2(1-x) which isn't a reflection?
Yes it is.
11. (Original post by langnerd)
I got y=log2(1-x) which isn't a reflection?
Seems like a pretty damn juicy reflection to me...

12. (Original post by langnerd)
I havent really ever been taught how to find the inverse of an exponential function but have been asked to find the inverse of f(x)= 1-2^x

Can anyone help?
This is how I did it:

let x=y:

Now you can take logs, natural logs or :

Now use the power rule of logarithms:

Rearrange for y:

That's how i'd do it anyway and I believe that's correct (checked on wolfram alpha) if you took log base 2 you could have a different result:

as
13. (Original post by L33t)
The methods thus far on here are wrong. In order to find an inverse function with an exponential you must end up with a log or ln somewhere! This is what allows reflection in y=x:

let x=y:

Now you can take logs, natural logs or :

Now use the power rule of logarithms:

Rearrange for y:

That's how i'd do it anyway and I believe that's correct (checked on wolfram alpha) if you took log base 2 you could have a different result.
That's what we all said, plus is the same curve as anyway.
14. (Original post by L33t)
...
Try not to solve other people's question's for them. Just point them in the right direction.
15. (Original post by RDKGames)
Seems like a pretty damn juicy reflection to me...

Ah thank you, Google isnt very good at drawing graphs it would seem!
16. (Original post by RDKGames)
Try not to solve other people's question's for them. Just point them in the right direction.
Calm down, I'm only trying to help, sorry I won't in future!
17. (Original post by langnerd)
Ah thank you, Google isnt very good at drawing graphs it would seem!

(Original post by L33t)
Calm down, I'm only trying to help, sorry I won't in future!
No problem, just saying
18. (Original post by L33t)
Yes! But for it to be an inverse function you have to swap x and y AND rearrange for the new y or it isn't correct. No marks in the exam for that.
Yeah but that's only because it says you need to find . If then sure .
19. (Original post by RDKGames)

No problem, just saying
I'm new to TSR so I don't know the etiquette yet you see, truly sorry.
20. (Original post by RDKGames)
Try not to solve other people's question's for them. Just point them in the right direction.
While I very much appreciate your efforts and admire your enthusiasm, I would also say the same thing. Eg

(Original post by RDKGames)
Prove that it now works for n=k+1

QED.

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