Question 1134064
Given that {{{3log(2,(x))}}} = y and {{{log(2,(4x))}}} = y + 4, where x and y are real numbers, find the values of x and y.
:
rewrite
{{{log(2,(x^3)) = y}}}
and
{{{log(2,(4x)) -4 = y}}}
y=y therefore
{{{log(2,(4x)) - 4}}} = {{{log(2,(x^3))}}}
rewrite to
{{{log(2,(4x)) - log(2,(x^3)) = 4}}}
which is
{{{log(2,((4x)/x^3)) = 4 }}}
cancel x
{{{log(2,(4/x^2)) = 4 }}}
the exponent equiv
{{{2^4 = 4/x^2}}} = {{{16 = 4/x^2}}}
{{{16x^2 = 4}}} = {{{x^2 = 4/16}}} = {{{x^2 = 1/4}}} = {{{x = sqrt(1/4)}}}
{{{x=1/2}}}
:
find y
{{{y = log(2,((1/2)^3))}}} = {{{y = log(2,((1/8)))}}}
exponent equiv
{{{2^y = 1/8}}}
{{{y = -3}}}
:
:
Check solutions in original 2nd equation: {{{log(2,(4x)) = y + 4}}},
{{{log(2,(4(1/2))) = -3 + 4}}},
{{{log(2,(2)) = 1}}},