Question 634223
  <pre><font face = "Tohoma" size = 3 color = "indigo"><b> 
Hi,
*[tex \large\ \ log_3(x-4)] + *[tex \large\ \ log_3(x-2)] =*[tex \large\ \ log_3(x-4)(x-2) ]= 1
 3^1 = (x-4)(x-2)
   x^2 - 6x + 5 = 0
  (x - 5)(x-1) = 0   
         x = 5     (x = 1  is an extraneous solution)
and 
*[tex \large\ \ log_3 (1*3) = 1]