Question 1114412
<br>
(1) {{{2^x*4^y=1}}}  -->  {{{2^x*2^(2y)=1}}}  -->  {{{2^(x+2y)=1}}}  -->  {{{x+2y=0}}}  -->  {{{x = -2y}}}<br>
(2) {{{4^(xy)=1/16}}}  -->  {{{xy = -2}}}<br>
Substituting (1) into (2),<br>
{{{-2y^2 = -2}}}
{{{y^2 = 1}}}
{{{y = 1}}} or {{{y = -1}}}<br>
(a) y=1 --> x = -2
(b) y = -1 -->  x = 2<br>
Answer: 2 solutions: (x,y) = (-2,1) or (x,y) = (2,-1)