Question 37643
1. {{{ sqrt(x) - 2 = 1 }}}
{{{ sqrt(x) = 3 }}}
now square both sides:
x = 9


2. {{{ sqrt(x^3) = 27 }}}
well you would square both sides then take the cube root of both sides. Alternatively since "raising to a power" and "rooting" are opposite processes, it doesn't actually matter the order we do this, so i shall find the cube root first:


{{{ sqrt(x) = root(3, 27) }}}
{{{ sqrt(x) = 3 }}}
--> x = 9

3. {{{ 3^(sqrt(x)^2) = 9 }}}
or... {{{ 3^(sqrt(x)^2) = 3^2 }}}
so by "symmetry" of this, we can say that {{{ sqrt(x^2) = 2 }}}
and also, {{{sqrt(x^2) }}} is just x anyway, so
x = 2


jon.