How to Linearize a Function without Using Logarithms

[mrwest09]

Homework Statement

Linearize the following model:
y=\alpha*x*e^{\beta*x}

Homework Equations

The only relevant equations I can think of are the laws of natural logarithms.

The Attempt at a Solution

I have tried to taking the ln of both sides however that leaves me with an equation that has two term with x in it.

ln(y)=ln(a)+ln(x)+Bx

I'm sure there has to be a simple solution but I can't visualize anything without running into the same problems.

 

[Mark44]

Homework Statement

Linearize the following model:
y=\alpha*x*e^{\beta*x}

Homework Equations

The only relevant equations I can think of are the laws of natural logarithms.

The Attempt at a Solution

I have tried to taking the ln of both sides however that leaves me with an equation that has two term with x in it.

ln(y)=ln(a)+ln(x)+Bx

I'm sure there has to be a simple solution but I can't visualize anything without running into the same problems.

It's possible that you're supposed to do this using a Maclaurin series representation for your function, and discard the x² and higher degree terms.

The Maclaurin series for e^\betax is
e^{\beta x} = 1 + \frac{\beta x}{1!} + \frac{(\beta x)^2}{2!} + ... + \frac{(\beta x)^n}{n!} + ...

Multiply the terms in this series by \alphax and then discard all terms in x² or higher.

EDIT:
On second thought, there's a simpler formula that is related to the above.

If x is "close to" 0, then f(x) \approx f(0) + x*f'(0). This gives you a first degree polynomial approximation to your function.

 

[Dickfore]

Homework Statement

Linearize the following model:
y=\alpha*x*e^{\beta*x}

Homework Equations

The only relevant equations I can think of are the laws of natural logarithms.

Correct:

<br /> \ln{y} = \ln{\alpha} + \ln{x} + \beta \, x<br />

So, \ln{y} - ln{x} is a linear model relative to the function x and you can use linear least squares fit to extract the value of the coefficients \ln{\alpha} (the intercept) and \beta (the slope).

 

[mrwest09]

Okay that makes some sense but if you were to fit that into your standard y=mx+b format for linear lines wouldn't your 'y' value depend on two variables? In this case wouldn't it not be linear?
This is how I am picturing the final equation:

<br /> <br /> \ln{y} - \ln{x} = \beta \, x + \ln{\alpha}<br /> <br />

y = m x + b

 

[Dickfore]

Yes. In:

<br /> \tilde{y} = m \tilde{x} + b<br />

we need to calculate:

<br /> \tilde{y} = \ln{y} - \ln{x}<br />

<br /> \tilde{x} = x<br />

and then:

<br /> m = \beta, \; b = \ln{\alpha}<br />