Inverse of exponential functions

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?

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?

So if you have $y=f(x)$ so here $y=1-2^x$ 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 $x=1-2^y$. Normally though they want the equation in the form $y=g(x)$ 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?

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.

I got y=log2(1-x) which isn't a reflection?

I got y=log2(1-x) which isn't a reflection?

I got y=log2(1-x) which isn't a reflection?

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?

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:

$y=1-2^x$

let x=y:

$x=1-2^y$

$x-1=-2^y$

$1-x=2^y$

Now you can take logs, natural logs or $log_2$:

$ln(1-x)=ln(2^y)$

Now use the power rule of logarithms:

$ln(1-x)=yln(2)$

Rearrange for y:

$\frac{ln(1-x)}{ln(2)}=y$

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.

...

Seems like a pretty damn juicy reflection to me...

Try not to solve other people's question's for them. Just point them in the right direction.

Ah thank you, Google isnt very good at drawing graphs it would seem!

Calm down, I'm only trying to help, sorry I won't in future!

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.

Try www.desmos.com/calculator instead



No problem, just saying

Try not to solve other people's question's for them. Just point them in the right direction.

Prove that it now works for n=k+1

$\displaystyle\sum_{r=1}^{k+1} r = \sum_{r=1}^{k} r + (k+1) = \frac{1}{2}k(k+1) + (k+1) = (k+1)[\frac{k}{2}+1] = \frac{1}{2}(k+1)(k+2)=\frac{k+1}{2}([k+1]+1)$

QED.