Question 854923
<pre><font face = "Tohoma" size = 3 color = "indigo"><b> 
Hi,
log2 (y+1) + log2 (3y-1) = 5 
     (y+1)(3y-1)) = 2^5
     3y^2 + 2y - 1 = 32
     3y^2 + 2y - 33 = 0
x = {{{(-2 +- sqrt( 400 ))/(6) }}} (Tossing out this negative root as Extraneous)
x ={{{(-2 +20)/(6) = 3 }}}