My understanding: Equation $(5)$ can be obtained when we let $e^x=\left({e^t}/{\theta}\right)^{\beta}$, which $x=\beta\,(t-\log\theta)$ , and substitute into $(4)$.
My question: Why equation $(7)$ is $\beta\,\left(Y-\log\theta\right)$ instead of $\beta\,(T-\log\theta)$ ? These lead to me not understanding the equation $(8)$ as well.
We know that the CDF of a complete Weibull Distribution as below :
\begin{equation}\text{Prob}\,(T<t)=1-e^{-\left({t}/{\theta}\right)^{\beta}}\tag{1}\label{eq:1}\end{equation}
The random variable $Y=\log T$ is called a log-Weibull variable. We have : \begin{align} \text{Prob}\,(Y < t) & = \text{Prob}\,(\log T < t) \tag{2}\label{eq: 2}\\ & = \text{Prob}\,(T < e^t)\tag{3}\label{eq: 3} \\ & = 1-e^{-\left({e^t}/{\theta}\right)^{\beta}} \tag{4}\label{eq: 4} \\ & = 1-e^{-e^{\beta(t-\log\,\theta)}} \tag{5}\label{eq: 5}\\ \end{align}
The random variable $X=\beta\,(Y-\log\,\theta)$ is called as a reduced log-Weibull variable. The distribution function $X$ is
\begin{align} F(x) & = \text{Prob}\,(X<x) \tag{6}\label{eq: 6}\\ & = \text{Prob}\,(\beta\,(Y-\log\,\theta)<x) \tag{7}\label{eq: 7} \\ & = 1-e^{-e^x} , \,\,\,-\infty<x<\infty\tag{8}\label{eq: 8} \\ \end{align}
Reference : https://www.jstor.org/stable/1267267