Exponential growth question

[MathsHelpPlz]

Not sure where this question belongs, but:

"N represents the number of cells at a time 't' minutes after the start of the growth, and the relationship between 'N' and 't' was thought to be modelled by N=a(b^t), where 'a' and 'b' are constants. Find the values of a and b to the nearest integer"

t	1.5	2.7	3.4	8.1	10
N	9	19	32	820	3100

I correctly found 'b' to be 2.
I drew a graph of Log2(N) against 't' and found the y-intercept to be 3.17, so, using Log2(N)=Log2(a) + tLog(b) I equated Log2(a)=3.17 and found 'a' as 2^3.17 ~ 9. However if I used the different method of letting 9=a(b^1.5) where b was found to be (19/9)^(1/2) via 19=9(b^1.2), I got 'a' as 9/(b^1.5) which is 3.53674416 ~ 4.

t	0	1.2	1.9	6.6	8.5
Log2(N)	3.17...	4.25...	5.00	9.68...	11.60...

I was wondering how come I get different results with the different methods?
Thank you for your time.

 

[pappus]

Not sure where this question belongs, but:

"N represents the number of cells at a time 't' minutes after the start of the growth, and the relationship between 'N' and 't' was thought to be modelled by N=a(b^t), where 'a' and 'b' are constants. Find the values of a and b to the nearest integer"

t	1.5	2.7	3.4	8.1	10
N	9	19	32	820	3100

I correctly found 'b' to be 2. <-- to be a little bit more exact: \(\displaystyle b \approx 2.008 \)
I drew a graph of Log2(N) against 't' and found the y-intercept to be 3.17, so, using Log2(N)=Log2(a) + tLog(b) I equated Log2(a)=3.17 and found 'a' as 2^3.17 ~ 9. However if I used the different method of letting 9=a(b^1.5) where b was found to be (19/9)^(1/2) via 19=9(b^1.2), I got 'a' as 9/(b^1.5) which is 3.53674416 ~ 4.

t		1.2	1.9	6.6	8.5
Log2(N)	3.17...	4.25...	5.00	9.68...	11.60...

I was wondering how come I get different results with the different methods?
Thank you for your time.

I don't know why you used this method to calculate the value of a ...(?)

If you use the given mathematical model:

\(\displaystyle N = a \cdot b^t \)

you can use any of the pairs of the original table to determine a. I'll take (8.1 , 820):

\(\displaystyle 820 = a \cdot 2.008^{8.1}~\implies~ a = \frac{820}{2.008^{8.1}} \approx 2.892\)

 

[MathsHelpPlz]

I don't know why you used this method to calculate the value of a ...(?)

If you use the given mathematical model:

\(\displaystyle N = a \cdot b^t \)

you can use any of the pairs of the original table to determine a. I'll take (8.1 , 820):

\(\displaystyle 820 = a \cdot 2.008^{8.1}~\implies~ a = \frac{820}{2.008^{8.1}} \approx 2.892\)

I think the data isn't completely exponential as I get different values of b for different bits of data I use. I know the algebraic way is easier, but I'm unsure why the graphial method isn't working and I might need to use it. The graph is Log2(N)=Log(a) + tLog(b), so the y-intercept should be Log(a), which equals the y-intercept of 3.17 from the Log2(N) table above. So Log2(a)=3.17 giving a=2^3.17 ~ 9

 

[pappus]

I think the data isn't completely exponential as I get different values of b for different bits of data I use. I know the algebraic way is easier, but I'm unsure why the graphial method isn't working and I might need to use it. The graph is Log2(N)=Log(a) + tLog(b), so the y-intercept should be Log(a), which equals the y-intercept of 3.17 from the Log2(N) table above. So Log2(a)=3.17 giving a=2^3.17 ~ 9

Since b = 2 (rounded of course!) you want to use the logarithm to the base 2.

I'm going to use the pair of data (3.4 , 32):

\(\displaystyle \displaystyle{\log_2(32)=\log_2(a)+3.4 \cdot \underbrace{\log_2(2)}_{=1}}\)

\(\displaystyle \displaystyle{5=\log_2(a)+3.4 }\)

\(\displaystyle \displaystyle{1.6=\log_2(a)~\implies~a=2^{1.6} \approx 3.03 }\)

Could it be that you have mixed logarithms to base 2 with logarithms to base 10?

 

[MathsHelpPlz]

Since b = 2 (rounded of course!) you want to use the logarithm to the base 2.

I'm going to use the pair of data (3.4 , 32):

\(\displaystyle \displaystyle{\log_2(32)=\log_2(a)+3.4 \cdot \underbrace{\log_2(2)}_{=1}}\)

\(\displaystyle \displaystyle{5=\log_2(a)+3.4 }\)

\(\displaystyle \displaystyle{1.6=\log_2(a)~\implies~a=2^{1.6} \approx 3.03 }\)

Could it be that you have mixed logarithms to base 2 with logarithms to base 10?

Ah I see what is wrong, it is because I am letting time 0 equal time 1.5, meaning the y-intercept is at 1.5 minutes away from a, so if I did 1.5 minutes as 1.5 minutes the y-intercept would be a.

 

[pappus]

ÿþ[ ÿþ[ ÿþ[

Could it be that your wheel-chair needs a drop of oil?

 

[HallsofIvy]

That's elvish isn't it?