I am trying to solve the following equations:
$0.5 = exp(-(3*c)^k)$ and $0.99 = exp(-(29*c)^k)$
I have used MATLAB to get the answers of $c = 0.21487$ and $k = 0.83471$ but I'd really like to know the method if anybody has any idea how to do it?
Many thanks!
taking the logarithm we get $-\ln\left(\frac{1}{2}\right)=(3c)^{k}$ and $-\ln(0.99)=(29c)^k$ and taking the logarithm again we have $$\ln\left(-\ln\left(\frac{1}{2}\right)\right)=k\ln(3c)$$ and $$\ln(-\ln(0.99))=k\ln(29c)$$ now you can divide both equations to get $c$ i got the following solutions $$c\approx 0.4056727050$$,$$k \approx -1.866120254$$