Two unknown variables in experiment equation

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Homework Help Overview

The discussion revolves around an experimental equation involving two unknown variables, A and d, in the context of a linear relationship derived from a logarithmic transformation of the equation. Participants are tasked with determining these variables based on a linear fit slope and y-intercept.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the relationship between the original equation and its logarithmic form, questioning how to manipulate the equation to isolate variables. There is discussion about the implications of the linear trend and how it relates to the parameters A and d.

Discussion Status

Some participants have made algebraic connections between the original equation and the linear form, suggesting that the logarithmic transformation can yield insights into the values of A and d. There is a recognition of the relationships between the coefficients in the linear equation and the parameters in the original equation, although no consensus on the final approach has been reached.

Contextual Notes

Participants are working under the constraints of an experimental setup and are required to derive values from given data points, specifically a slope and y-intercept from a linear fit.

smashyash
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Homework Statement



I'm given an experimental equation:

y = sqrt[A^2*exp(x/d)]

(with a linear trend of ln(y^2) versus x)

I am suppose to determine the values of first A, then d given a linear fit slope and a y intercept.

Homework Equations



y = mx + b

The Attempt at a Solution



At first, I simply took the y intercept value and plugged that into y and plugged 0 in for x. I thought this would give me A since no matter what the value of d at that point, the exp will be exp(0) = 1. But this is not the case and I'm not sure how to use the given slope. Should I try setting the y = mx + b equal to the equation and solve for the variables??
 
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The linear trend means:

ln(y^2) = mx + b

Btw, what would ln(y^2) evaluate to?
 
I'm really not sure...

I don't really see how ln(y^2) relates to the equation, there's no way to manipulate the equation to get just that on one side..
 
You know y = sqrt[A^2*exp(x/d)], so what is y^2? How about ln(y^2)?

BTW it's useful to know that ln(a*b)=ln a + ln b
 
Ok, so here's the algebra I have:

y^2 = A^2*exp(x/d)

ln(y^2) = ln( A^2 * exp(x/d) )

ln(y^2) = ln(A^2) + x/d

So if this is true, then is ln(A^2) = b and 1/d = m?
y = mx + b
 
Yes, that's exactly right. You're supposed to determine the values of A and d, though, so just invert the equations you already have.
 
great! thanks so much! :)
 

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