For Weibull distribution,  
S(t) = $e^{-(\lambda * e^(x * \beta)*t)^\rho}$

 "$^{(1/rho)}$" will be only for log(v)


so, I modified like this
  

    Tlat <- - log(v)^(1 / rho) / (lambda * exp(x * beta))

if rho = 1, result will be same.