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^$\beta$x 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 $\alpha$x 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:

$$\ln{y} = \ln{\alpha} + \ln{x} + \beta \, x$$

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:

$$\ln{y} - \ln{x} = \beta \, x + \ln{\alpha}$$

y = m x + b

 

[Dickfore]

Yes. In:

$$\tilde{y} = m \tilde{x} + b$$

we need to calculate:

$$\tilde{y} = \ln{y} - \ln{x}$$

$$\tilde{x} = x$$

and then:

$$m = \beta, \; b = \ln{\alpha}$$