theorem Th2:
  a>0 implies exp_R.(-x*log(number_e,a))=a #R (-x)
proof
A1: number_e<>1 by TAYLOR_1:11;
  assume
A2: a>0;
  exp_R.(-x*log(number_e,a)) =exp_R.((-x)*log(number_e,a))
    .=exp_R.(log(number_e,a to_power (-x))) by A2,A1,POWER:55,TAYLOR_1:11
    .=exp_R.(log(number_e,a #R (-x))) by A2,POWER:def 2
    .=a #R (-x) by A2,PREPOWER:81,TAYLOR_1:15;
  hence thesis;
end;
