I have to rearrange the following, making p the subject of the equation:

ln(p) = β0 + β1x

I have got P = e^(β0 + β1x) Is this correct

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- Feb 28th 2009, 08:06 AMOrangeobe[SOLVED] Natural Logarithm
I have to rearrange the following, making p the subject of the equation:

ln(p) = β0 + β1x

I have got P = e^(β0 + β1x) Is this correct - Feb 28th 2009, 08:13 AMJhevon
- Feb 28th 2009, 08:23 AMe^(i*pi)
Yes it is correct but you can use the rules of exponentials to say that $\displaystyle e^{(\beta0 + \beta1x)} = e^\beta0 * e^{(\beta1x)}$

$\displaystyle If we say that \alpha = e^{\beta0}$ then we get

$\displaystyle p = \alpha e^{(\beta1x)}$

This assumes $\displaystyle \beta0$ is a constant, if it is not constant than ignore this whole post