Please help me figure out how this exponential model was determined from the table.

[endoplasmicreticulum]

Can someone please explain how this is the answer? I am simply confused on how they arrived to that answer for the exponential model. Because, the answer I got was f(x) = 16.634(0.250)^x

I thought that to find the exponential model, you were supposed to recognize the pattern of y, (which is multiplying by 0.250, no?) and then, to find the "a" in y = ab^x, you would just insert the y value where the input = 0 (in this case, 16.634 ?)

Any help would be greatly appreciated, thanks.

 

[mario99]

Can someone please explain how this is the answer? I am simply confused on how they arrived to that answer for the exponential model. Because, the answer I got was f(x) = 16.634(0.250)^x

I thought that to find the exponential model, you were supposed to recognize the pattern of y, (which is multiplying by 0.250, no?) and then, to find the "a" in y = ab^x, you would just insert the y value where the input = 0 (in this case, 16.634 ?)

Any help would be greatly appreciated, thanks.

You should do the years like this:

[imath]1940: x = 0 \rightarrow y = 16.634[/imath]
[imath]1965: x = 25 \rightarrow y = 4.091[/imath]
[imath]1990: x = 50 \rightarrow y = 1.023[/imath]
[imath]2015: x = 75 \rightarrow y = 0.256[/imath]

 

[endoplasmicreticulum]

You should do the years like this:

[imath]1940: x = 0 \rightarrow y = 16.634[/imath]
[imath]1965: x = 25 \rightarrow y = 4.091[/imath]
[imath]1990: x = 50 \rightarrow y = 1.023[/imath]
[imath]2015: x = 75 \rightarrow y = 0.256[/imath]

That's right, I've figured out how to solve the problem after some time. Thanks for your input.

 

[Dr.Peterson]

Can someone please explain how this is the answer? I am simply confused on how they arrived to that answer for the exponential model. Because, the answer I got was f(x) = 16.634(0.250)^x

I thought that to find the exponential model, you were supposed to recognize the pattern of y, (which is multiplying by 0.250, no?) and then, to find the "a" in y = ab^x, you would just insert the y value where the input = 0 (in this case, 16.634 ?)

Any help would be greatly appreciated, thanks.

If you do this by hand, just recognizing the approximate multiplier as you say, you need to take into account that it is multiplied by 0.250 every 25 years, not every year, in order to get the correct base. I suppose that is the error you recognized.

I don't know how you'd be expected to find the initial value in the formula (namely the 16.549 as opposed to 16.634).

It is what Excel gives me when I ask for a trend line:

​

The exponential part can be converted to your form: [imath]e^{-0.056}=0.9455[/imath].

To correct your answer, [imath]f(x) = 16.634(0.250)^x[/imath], you need to change it to [imath]f(x) = 16.634(0.250)^{x/25}=16.634(0.250^{1/25})^x=16.634(0.946)^x[/imath].

 

[mario99]

That's right, I've figured out how to solve the problem after some time. Thanks for your input.

You are welcome.

I don't know how you'd be expected to find the initial value in the formula (namely the 16.549 as opposed to 16.634).

I am guessing that the OP has used that value directly from the table or he has used wrong values for [imath]x[/imath].

I know that it sounds crazy, but I have done all the calculations by hand (not pure hand). First, I have calculated [imath]\ln y[/imath] as if it were a linear regression.

[imath]\ln y = mx + b[/imath]


[imath]\displaystyle m = \frac{n\sum (x\ln y) - (\sum x)(\sum \ln y)}{n\sum x^2 - (\sum x)^2} \approx -0.05563252[/imath]


[imath]\displaystyle b = \frac{\sum \ln y - m\sum x}{n} \approx 2.80632[/imath]

I was just lazy to plug in every number. I got this:

[imath]\ln y = -0.05563252x + 2.80632[/imath]

Playing around to get the required form.

[imath]f(x) = y = (e^{-0.05563252})^{x} e^{2.80632} = (0.946)^{x}16.549 = 16.549(0.946)^{x}[/imath]