Question 29426
i think you mean {{{ ln( log_x(2)) = -1 }}} where log_x means log to base x.


So, {{{ log_x(2) = e^(-1) }}}
{{{ log_x(2) = 1/e }}}


Now convert the log to base 10: --> future wording of "log" means to base 10:
{{{ log(2)/log(x) = 1/e }}}
{{{ elog(2) = log(x) }}}
{{{ 0.818284367 = log(x) }}}


and so, x = 6.580885991


jon.