# weibull parameter

• Oct 30th 2008, 12:21 PM
weibull parameter
Hi All..

I wanted to know how do you make the Weibull equation below into the form y=mx+c. The solution to this has the logs taken twice..

$P_f= 1 - exp\left[-\left(\frac{\sigma}{\sigma{_0}}\right)^m\right]$

Thank you
• Oct 30th 2008, 01:38 PM
mr fantastic
Quote:

Hi All..

I wanted to know how do you make the Weibull equation below into the form y=mx+c. The solution to this has the logs taken twice..

$P_f= 1 - exp\left[-\left(\frac{\sigma}{\sigma{_0}}\right)^m\right]$

Thank you

This is the cdf of the Weibull distribution. It's not clear to me what you're actually trying to do with it. Post the whole question, exactly as it's written.
• Oct 30th 2008, 01:54 PM
Hi.. thanks for your reply.. there is no question just from notes where the weibull is in the form y=mx+c.. just wanted to know the process on how it is derived?

I will put the solution..
$\ln\ln\left(\frac{1}{1-P_f}\right) = m\ln\sigma - m\ln\sigma_0$

Thanks
• Oct 30th 2008, 02:09 PM
Sean12345

We have
$P_f=1-exp\left[-\left(\frac{\sigma}{\sigma{_0}}\right)^m\right]$

Note that we can re-write this as

$exp\left[-\left(\frac{\sigma}{\sigma{_0}}\right)^m\right]=1-P_f$

Taking logs to the base $e$ of both sides gives

$-\left(\frac{\sigma}{\sigma{_0}}\right)^m=\ln(1-P_f)$

Take the minus over to make it easier to deal with....

$\left(\frac{\sigma}{\sigma{_0}}\right)^m=-\ln(1-P_f)$

Now we know that $a\ln(b)\equiv\ln(b^a)$ thus we are left with

$\left(\frac{\sigma}{\sigma{_0}}\right)^m=\ln\left( \frac{1}{1-P_f}\right)$

Taking logs of base $e$ again and using that same principle leads to

$
m\ln\left[\left(\frac{\sigma}{\sigma{_0}}\right)\right]=\ln\left[\ln\left(\frac{1}{1-P_f}\right)\right]
$

Using the fact that $\ln\left(\frac{a}{b}\right)\equiv \ln(a)-\ln(b)$ gives the required result.

Hope this helps.
• Oct 30th 2008, 02:19 PM