SOLUTION: How do I solve this problem: 4^1996+4^1995+4^1994+4^1993=45(2^x)? My calculator says infinity. I tried factoring it: 4^1993(85)=45(2^x), but my calculator still says infinity.

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: How do I solve this problem: 4^1996+4^1995+4^1994+4^1993=45(2^x)? My calculator says infinity. I tried factoring it: 4^1993(85)=45(2^x), but my calculator still says infinity.      Log On


   

Question 3741: How do I solve this problem: 4^1996+4^1995+4^1994+4^1993=45(2^x)? My calculator says infinity.
I tried factoring it: 4^1993(85)=45(2^x), but my calculator still says infinity.
Answer by khwang(438) About Me  (Show Source):
You can put this solution on YOUR website!
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