2^5x*16^1-x=4^x-3
2007-08-16 13:58:32 · 3 answers · asked by patrice cadet5 1 in Science & Mathematics ➔ Mathematics
2^5x*16^1-x=4^x-3
2^5x * 2^4(1-x) = 2^2(x-3)
2^(5x+4-4x) = 2^(2x-6)
2^(x+4) = 2^(2x-6)
x+4 = 2x-6
x=10
2007-08-16 14:15:21 · answer #1 · answered by ironduke8159 7 · 0⤊ 0⤋
Change the bases to 2. So it looks like this:
2^(5x) * 2^(4*(1-x)) = 2^(2(x-3))
Simplified, it is:
2^(5x) * 2^(4 -4x) = 2^ (2x- 6)
(x^m)(x^n)= X^(m + n), so:
2^(4+x) = 2^ (2x-6)
The bases are the same, so you can ignore it (you can prove this if you log the entire problem and then divide by log 2's on both sides).
4+x = 2x-6 is simple enough to solve, so I'll leave that part to you. :)
2007-08-16 21:16:26 · answer #2 · answered by Jess 2 · 0⤊ 0⤋
2^(5x) 16^(1-x) =4^(x-3)
2^(5x) 2^{4(1-x)} =2{2(x-3)}
2^(5x) 2^(4-4x) divided by 2^(2x-6) =1
2^(5x) 2^(4-4x) 2^(-(2x-6)) =2^0
5x + (4-4x)-(2x-6) =0
5x+4-4x-2x+6=0
-x+10=0-x=-10
x=10
Check:
2^50 (2^4)^-9 = 2^(2(7))
2^50/2^36 = 2^14
2^14 = 2^14
A lot easier with pencil & paper!
2007-08-16 21:32:17 · answer #3 · answered by Grampedo 7 · 1⤊ 0⤋