<PRE><B>Question 98517</B>
2log y = log 2 + log x , 2^y=4x

-----------------------
log(y^2/x) = log(2)
---------------------------------
1st: y^2 = 2x
------------------
2nd: 2^y = 4x
--------------
Multiply 1st by 2 to get:
2y^2=4x
------------
Therefore 2y^2 = 2^y
y^2 = 2^(y-1)
--------------
Solution: y = 1
Substitute into 2logy=log2+logx
0 = log2 +logx
0=log(2x)
2x = 1
x = 1/2
----------------
Check x=1/2 and y=1 in 2^y=4x
2^1 = 4(1/2)
2 = 2
===========
Cheers,
Stan H.
===============
Cheers,
Stan H.</PRE>