Question 893221

If {{{y=2^x}}}
express the following in terms of y:
a) {{{2^(2x+1)}}}
b) {{{2^(1-x)}}}

I tried doing it a few ways, could exactly figure out what the questions want.
The answers are:
a) {{{2y^2}}}
b) {{{2/y}}}
<pre>
{{{2^(2x+1)}}}______{{{2^(2x) * 2^1}}}______{{{2^x * 2^x * 2^1}}}
Since {{{y = 2^x}}}, then {{{2^x * 2^x * 2^1}}} becomes: y * y * 2, or {{{highlight_green(2y^2)}}}

{{{2^(1 - x)}}}____{{{2^1 * 2^(- x)}}}____{{{2^1 * (2^(-1))^x}}}____{{{2 * (1/2)^x}}}____{{{2 * (1/2^x)}}}____{{{2/2^x}}} 
Since {{{y = 2^x}}}, then {{{2/2^x}}} becomes: {{{highlight_green(2/y)}}}