Question 464677
  <pre><font face = "Tohoma" size = 4 color = "indigo"><b> 
Hi,
log4[x]+log4[x-63]=3
log4[x(x-63)] = 3
*[tex \LARGE \ \ \ \ \ \ \ \ \ \ log_b(x) \ = \ y \ \ \Rightarrow\ \ b^y = x]
 4^3 = x(x-63)
  x^2 -63x - 64 = 0
factoring:
 (x-64)(x+1) = 0
  (x-64)= 0  x = 64
  (x+1) = 0  x = -1 Extraneous solution
  64 is the real solutions for x