1. ## 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..

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

Thank you

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..

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

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.

3. 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..
$\displaystyle \ln\ln\left(\frac{1}{1-P_f}\right) = m\ln\sigma - m\ln\sigma_0$

Thanks

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

Note that we can re-write this as

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

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

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

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

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

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

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

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

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

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

Hope this helps.

5. ## thanks

Thank you.. that's exactly what I needed..You made it very clear line by line..