Obtaining a Future Numbers Increase With Exponent From Past Increase

[Point Conception]

TL;DR

With exponent from past numbers increase, apply to obtain future increase

Suppose on day-one a number is 15 then on day-twenty the number has increased to 200. Now I want to find out what that increasing number could be on day-forty by using the exponent derived from the day- one to day-twenty increase ; x(log15) = log 200 .
x = 2.301/1.176 = 1.956. So now on day twenty can I predict what this increasing number will be on day-forty applying exponent (1.956) to the count on day twenty (200) 200^1.956 = 31,682

 

[Mark44]

Summary:: With exponent from past numbers increase, apply to obtain future increase

Suppose on day-one a number is 15 then on day-twenty the number has increased to 200. Now I want to find out what that increasing number could be on day-forty by using the exponent derived from the day- one to day-twenty increase ; x(log15) = log 200 .

Are you assuming that the numbers grow exponentially? Is there some reason that exponential growth is reasonable? Your equation just above is equivalent to $15^x = 200$.

For exponential growth, the equation is typically given as $N(t) = N_0e^t$.

x = 2.301/1.176 = 1.956. So now on day twenty can I predict what this increasing number will be on day-forty applying exponent (1.956) to the count on day twenty (200) 200^1.956 = 31,682

 

[Point Conception]

Are you assuming that the numbers grow exponentially? Is there some reason that exponential growth is reasonable? Your equation just above is equivalent to $15^x = 200$.

For exponential growth, the equation is typically given as $N(t) = N_0e^t$.

Yes assuming exponential growth (flu cases) but just for pure math is this correct. Yes 15^1.956 = 200, as shown and obtained in post, So now can I apply that determined exponent to 200 , the count reached on day twenty : 200^1.956 = 31,682? to obtain the count on day forty,3,682. With the typical exponential growth equation would that be N(t) = 200 e^20 or would that be N(t)=15e^40 ?

 

[mfb]

Your calculation doesn't lead to an exponential increase.
Simple cross checks:

• If the number at day 1 would be 0, your calculation would fail, but an exponential function has a value of 1 somewhere.
• You didn't use the days anywhere, but clearly it matters when you have 15 and 200 cases and when you want to know the result.
• If you double both initial numbers then all other numbers should double, but the rate of increase shouldn't change. It would in your calculation.

All three of them fail.

Start from scratch. Use the equation for an exponential function, determine the two free parameters using the two data points you have, then calculate the unknown value based on that.

 

[Mark44]

Start from scratch. Use the equation for an exponential function, determine the two free parameters using the two data points you have, then calculate the unknown value based on that.

Strongly agree...

 

[Point Conception]

15 to 200 in 20 days. N(t) = Noe^kt, e^20k =13.33. 20k = ln13. 3 k =.13 So at 40 days N= 15e^40(.13) = 2719

 

[Point Conception]

And sure above is correct for N(40) since N(20) agreed: N(20) = 15e^20(.13) = 201

 

[mfb]

It's hard to read the calculations with that formatting. Line breaks would help already, but we have LaTeX support

15 to 200 in 20 days.

19 days?

Be careful with rounding exponents, you quickly get large errors that way. If you assume 20 days between the original points then at 40 days you should get 2667 instead of 2719.

 

[Svein]

If you want to fit an exponential curve to a time series:

1. Take the log of all the y-values.
2. Do a linear regression of of the log values against the time values
3. Now you have a best fit of the type log(y) = A⋅t +B
4. Take the exponential: y=e^{At+B}