Help with finding equation through logarithmic linearization

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Waddlingduck

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hello!



I have been given a task where i should use logarithmic linearization to find a functions exponent with the base of x. So i need to make an exponential function which corresponds to the values of the table below.
The function should look something like y = C + axn, and after the logarithmic linearization it looks like ln(y-c) = ln(a) + n*ln(x) which is a linear function. my teacher told me that the "n" is the incline of the funtion and is therefore the exponent I'm searching for.
the thing is, I dont know how to do logarithmic linearization when i have a zero value in the table. I've tried to simply ignore that value but that gives me the wrong function that doesn't correspond to the table.


Any help? my teacher told me that this exercise is extremely easy if you know how to do it, but I have been here trying for about 5 hours.


x 0,0 1,0 2,0 5,0 7,0 10,0
y 3,0 5,0 8,7 25,4 40,0 66,2
 
Edit: I think it's called log-linear analysis in english, but i have a hard time applying in to this table.
 
Waddlingduck said:
hello!



I have been given a task where i should use logarithmic linearization to find a functions exponent with the base of x. So i need to make an exponential function which corresponds to the values of the table below.
The function should look something like y = C + axn, and after the logarithmic linearization it looks like ln(y-c) = ln(a) + n*ln(x) which is a linear function. my teacher told me that the "n" is the incline of the funtion and is therefore the exponent I'm searching for.
the thing is, I dont know how to do logarithmic linearization when i have a zero value in the table. I've tried to simply ignore that value but that gives me the wrong function that doesn't correspond to the table.


Any help? my teacher told me that this exercise is extremely easy if you know how to do it, but I have been here trying for about 5 hours.


x 0,0 1,0 2,0 5,0 7,0 10,0
y 3,0 5,0 8,7 25,4 40,0 66,2
Click to expand...
You are correct that the log of 0 is not a real number, but \(\displaystyle a * 0^n\) is what real number?

Do you see now how to compute C?

Consider the case when x is 1. What is the value of \(\displaystyle a * 1^n\)?

if you found C correctly, you should now be able to find the numerical value of a. Can you do that?

If so, you have only one unknown remaining, namely n. Now use logs to find n.

Do as much as you can, then show us your work as far as you can go or show us your answer with supporting work, and we can help you finish or confirm your work.
 
JeffM said:
You are correct that the log of 0 is not a real number, but \(\displaystyle a * 0^n\) is what real number?

Do you see now how to compute C?

Consider the case when x is 1. What is the value of \(\displaystyle a * 1^n\)?

if you found C correctly, you should now be able to find the numerical value of a. Can you do that?

If so, you have only one unknown remaining, namely n. Now use logs to find n.

Do as much as you can, then show us your work as far as you can go or show us your answer with supporting work, and we can help you finish or confirm your work.
Click to expand...

I have done some calculations now.

y= C + a*x^n

3 = C proven when x = 0

y = 3 + a*x^n

5 = 3 + a*1^n makes a = 2

8.7 = 3 + 2*2^n

ln((8.7-3)/ (5/3)) = n*ln(2)

n = 0.354

so the function should be y = 3 + 2*x^(0.35), but this function grows to slowly for the table. Can you see anything wrong with my calculations?
 
Last edited:
Waddlingduck said:
I have done some calculations now.

y= C + a*x^n

3 = C proven when x = 0 Yes. Now if these were empirical observations, you might want to say \(\displaystyle C\approx 3.\)

y = 3 + a*x^n

5 = 3 + a*1^n makes a = 2 Yes, but see above. So

\(\displaystyle y = 3 + 2x^n \implies y - 3 = 2x^n \implies ln(y - 3) = ln(2) + nln(x) \implies n = \dfrac{ln(y - 3) - ln(2)}{ln(x)}.\)

8.7 = 3 + 2*2^n

ln((8.7-3)/ (5/3)) = n*ln(2) Where did the 5/3 come from?

n = 0.354

so the function should be y = 3 + 2*x^(0.35), but this function grows to slowly for the table. Can you see anything wrong with my calculations?
Click to expand...
When you tested the values greater than x = 2 and got a mismatch, one possibility is that you made an error in your arithmetic.

\(\displaystyle n = \dfrac{ln(8.7 - 3) - ln(2)}{ln(2} = \dfrac{ln(5.7/2)}{ln(2)} = \dfrac{ln(2.85)}{ln(2)} \approx 1.51 \approx 1.5 = \dfrac{3}{2}.\)

So \(\displaystyle y \approx 3 + 2\sqrt{x^3}.\) Let's see how that works on the other observations.

\(\displaystyle 3 + 2\sqrt{5^3} \approx 3 + 2 * 11.2 = 3 + 22.4 = 25.4.\) Looks good.

\(\displaystyle 3 + 2\sqrt{7^3} \approx 3 + 2 * 18.5 = 3 + 37.0 = 40.0.\) Looks good.

\(\displaystyle 3 + 2\sqrt{10^3} \approx 3 + 2 * 31.6 = 3 + 63.2 = 66.2.\) Looks good.
 
lookagain said:
8.7 = 3 + 2*2^n

8.7 - 3 = 2*2^n

4.7 = 2*2^n

4.7/2 = 2^n

2.35 = 2^n

I'm using the "log" (base 10) button on my calculator, even though the "ln" button is okay, too.

log(2.35) = log(2^n)

log(2.35) = n*log(2)

[log(2.35)]/[log(2)] = n

n = 1.23266 \(\displaystyle \ \ \ when \ \ rounded \ \ to \ \ five \ \ decimal \ \ places \)




Waddlngduck, why did you use commas in your lists of x- and y-values for the decimal numbers,
but then you used decimal points (periods) for the decimal numbers later in your post?
Click to expand...

Sorry, i copied the table from the document i got the exercise from, which contained commas. I hope it didn't cause to much confusion.

so the function then becomes y = 3 + 2*x^(1.23266), and it doesn't really correspond to the table on the later values. should this function be considered correct for the table anyway?
 
lookagain said:
8.7 = 3 + 2*2^n

8.7 - 3 = 2*2^n

4.7 = 2*2^n According to my arithmetic, 8.7 - 3 = 5.7, not 4.7.

4.7/2 = 2^n

2.35 = 2^n

I'm using the "log" (base 10) button on my calculator, even though the "ln" button is okay, too.

log(2.35) = log(2^n)

log(2.35) = n*log(2)

[log(2.35)]/[log(2)] = n

n = 1.23266 \(\displaystyle \ \ \ when \ \ rounded \ \ to \ \ five \ \ decimal \ \ places \)




Waddlngduck, why did you use commas in your lists of x- and y-values for the decimal numbers,
but then you used decimal points (periods) for the decimal numbers later in your post?
Click to expand...
.
 
Waddlingduck said:
Sorry, i copied the table from the document i got the exercise from, which contained commas. I hope it didn't cause to much confusion.

so the function then becomes y = 3 + 2*x^(1.23266), and it doesn't really correspond to the table on the later values. should this function be considered correct for the table anyway?
Click to expand...
Lookagain made an error (easy enough to do). Check out my second post.
 
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