Question 3741
 You had better write 4^1993(85)=45(2^x) as
  85*4^1993=45(2^x), 
  
 Divide both sides by 5, and use 4=2^2, we have

 17*(2^2)^1993 = 9*2^x
  
 So, 17*2^3986 = 9*2^x
 
 And hence, 2^(x-3986) = 17/9
 
 Apply log2 (base 2) on both sides,(or use def of log), we get

 x-3986 =log2(17/9) = log(17/9)/log2 (base 10) ~ 0.9175

 [Note,this value is OK since 17/9 is close to 2 and log2 2 = 1]

 Thus,  x = 3986.9175

 By the way, there is no any reason that you tried to get the value
 4^1993 by using calculator. Clearly, it must be  +OO,because it is
 a huge number that no common calculator can handle it.

 My point is "Don't rely too much on calculator." Correct idea in math  
 is much more important than the nonsense and boring calculations.

 Kenny