Question 942780
<pre>
   a.) log sub2(log sub4{log sub3[log sub2 x]}) = -1
   b.) log sub2(x+4) + log sub2(x-2) - log sub2(x-6) = 0

a) PAINSTAKINGLY LONG solution by the other person.
   {{{log (2, (log (4, (log (3, (log (2, (x)))))))) = - 1}}}

b) Other person's solutions are EXTRANEOUS, and therefore, UNACCEPTABLE!!!
   b.) log sub2(x+4) + log sub2(x-2) - log sub2(x-6) = 0
       {{{log (2, (x + 4)) + log (2, (x - 2)) - log (2, (x - 6)) = 0}}}
       
        Smallest log: (x - 6), so x - 6 > 0. Therefore, x > 6.
        We then have: {{{log (2, (x + 4)) + log (2, (x - 2)) - log (2, (x - 6)) = 0}}}, <font color = red><font size = 4><b>with x > 6.</font></font></b>

The 2 solutions the other person got, x = - 2, and x = 1 are < 6, NOT > 6, obviously.
This makes them EXTRANEOUS, thereby leading to this equation having NO SOLUTIONS!</pre>