hey there,
(a) we know 2^x = 1/root 2.
taking logs of both sides gives
ln 2^x = ln [1/root2]
=> x ln 2 = ln 1 - ln [root 2]
= 0 - ln [root 2]
hence, x = -ln [ root 2 ] / ln 2 .
You may wish to chnage this a little more, but that answer is an exact value of x.
(b) first, lets calculate y .
Again taking logs,
ln 2^y = ln (4root 2)
=> y ln 2 = ln (4 root 2)
=> y = ln (4root2) / ln 2
Now lets consider y - x .
y - x = ln (4 root 2)/ln 2 - ( - ln (root2))/ ln 2
= [ln (4root2) + ln (root2)] / ln 2
= [ ln(4root2 * root2)]/ ln 2
= ln 8 / ln 2
so 2^ y-x = 2^ (ln 8/ ln 2)
I think this is correct. If there is an error let me know. All I've done is just use the laws of logs for the steps anyway !!!
Hope this helps