How to invert a simple exponential growth formula

[bakatu]

I think this is simple but my math skills are limited. I have a basic exponential growth formula: y=x*(1-p)ⁿ and I have y and x and values (0-11) and I need value of p. Then when I slove p, I have to calculate y with different values of n. It's easy for me to do that but than I get y values that decreases fast and than slow like on this graph:

But i want to decrease slower and than faster like on this graph:


I guess that formula needs somehow needs to be reversed, but I don't know how. Thanks in advance

 

[HallsofIvy]

You have \(\displaystyle y= x(1- p)^n\) and you want to solve for p?

That's fairly straight forward. You just "unpeel" p. First get rid of that "x" by dividing both sides by x: \(\displaystyle \frac{y}{x}= (1- p)^n\). Now get rid of that "power of n" by taking the nth root of both sides: \(\displaystyle \left(\frac{y}{x}\right)^{1/n}= 1- p\).

Can you finish it yourself?

 

[bakatu]

You have \(\displaystyle y= x(1- p)^n\) and you want to solve for p?

That's fairly straight forward. You just "unpeel" p. First get rid of that "x" by dividing both sides by x: \(\displaystyle \frac{y}{x}= (1- p)^n\). Now get rid of that "power of n" by taking the nth root of both sides: \(\displaystyle \left(\frac{y}{x}\right)^{1/n}= 1- p\).

Can you finish it yourself?

Thanks for the answer but I knew how to get that p. What I don't know is how to make the decay logarithmic.
For example I have:

y= 0.0089 and x= 0.0588 n=11

if use your formula I get:

0.0089=0.0588*(1-p)¹¹

p=0.15



BUT whith this kind of exponential decay I get for each n I get y that decreses faster and than slower. like this
But I want decrease at the beginig to be slower, and than faster like this
I thik that is called logharitmic decay.

I think that your formula somehow needs to be reversed so the big decresese donesn't happenes at the beging (between y1 and y2, y2 and y3 etc.)) but at the end of the last n values (between y9 and y10, y10 and y11).

I hope you can undestand what I need.

 

[Ishuda]

Substitute a for x and then what it seems to me you are saying is that you want some curve y = a (1-p)^x to be concave up instead of concave down.
y' = ln(1-p) y
y'' = [ln(1-p)]² y
So y'' is always positive if y is positive, i.e. a is positive, and you can not have a concave down curve.

So, lets look at
y = A - a (1-p)^x
with
y'' = -a [ln(1-p)]² (1-p)^x