Question 704768
{{{4=32^(1-2x) }}}
take log base 32 of both sides to get:
{{{log(32,4) = log(32,32^(1-2x)) }}}
on the right, it just reduces to:
{{{log(32,4) = (1-2x) }}}
apply "change of base" to the left:
{{{log(4)/log(32) = 1-2x }}}
{{{log(4)/log(32) - 1 = -2x }}}
multiplying both sides by -1:
{{{ -log(4)/log(32) + 1 = 2x }}}
or
{{{ 1-log(4)/log(32) = 2x }}}
dividing both sides by 2:
{{{ (1-log(4)/log(32))/2 = x }}}
{{{ (1-0.4)/2 = x }}}
{{{ (0.6)/2 = x }}}
{{{ 0.3 = x }}}
to four significant places:
{{{ 0.3000 = x }}}