1 thought on “Page<br />* Find value of x, y<br />(32)^x ÷2^y+1=1,16^4-x/2 -8^y=0<br />”
Answer:
Notice that y≥0, y≠1 and x≠0. If (x,y) is a solution then (−x,y) is a solution too. Hence W.L.O.G x>0.
It seems that x and y cannot be too far from each other, otherwise LHS would become larger than RHS. We’ll find out this. Suppose that z=|x−y|. Notice that z≠0. Now, we have
z2(x+y)2=|x−y|2(x+y)2=(x2−y2)2=16y+1
And since y≥0, y≠1 and x>0, then
z2=16y+1(x+y)2<4(1+y)2(1+y)2=4 ⟹ z<2
Therefore z=1. It follows that x=y±1 and equation turns into a simple quadratic for y.
Answer:
Notice that y≥0, y≠1 and x≠0. If (x,y) is a solution then (−x,y) is a solution too. Hence W.L.O.G x>0.
It seems that x and y cannot be too far from each other, otherwise LHS would become larger than RHS. We’ll find out this. Suppose that z=|x−y|. Notice that z≠0. Now, we have
z2(x+y)2=|x−y|2(x+y)2=(x2−y2)2=16y+1
And since y≥0, y≠1 and x>0, then
z2=16y+1(x+y)2<4(1+y)2(1+y)2=4 ⟹ z<2
Therefore z=1. It follows that x=y±1 and equation turns into a simple quadratic for y.
Step-by-step explanation:
I hope this helps u