solve for x : (2^(x+2))/ 16 = (4^(x-1))/ (2^(2x+1).
2007-07-30 18:20:42 · 4 answers · asked by Allison 1 in Science & Mathematics ➔ Mathematics
I'm sorry everyone the denominator on the LHS is suppose to read 16^x.
2007-07-30 19:02:25 · update #1
2^(x + 2) / 16^x = 4^(x - 1) / 2^(2x + 1)
2^(3x + 3) = 4^(x - 1) 16^x
2^(3x + 3) = 4^(x - 1) 4^(2x)
2^(3x + 3) = 4^(3x - 1)
2^(3x + 3) = (2²)^(3x - 1)
2^(3x + 3) = 2^(6x - 2)
3x + 3 = 6x - 2
3x = 5
x = 5 / 3
2007-07-30 20:00:52 · answer #1 · answered by Como 7 · 1⤊ 0⤋
(2^(x+2))/ 16^x = (4^(x-1))/ (2^(2x+1)
2^(x+2) / 2^(4x) = 2^(2x-2) / 2^(2x+1)
2^(x+2-4x) = 2^(2x-2-2x-1)
x+2-4x = 2x-2-2x-1
2-3x = -3
3x = 5
x = 5/3
Check:
LHS
= 2^(5/3 + 2)/16^(5/3)
= 2^(11/3) / 2^(20/3)
= 2^(-9/3)
RHS
= 4^((5/3)-1)/2^((10/3)+1)
= 4^(2/3) / 2^(13/3)
= 2^(4/3) / 2^(13/3)
= 2^(4/3 - 13/3)
= 2^(-9/3)
LHS = RHS
2007-07-31 01:24:26 · answer #2 · answered by gudspeling 7 · 0⤊ 0⤋
2^(x+2) / 16 = 4^(x-1) / 2^(2x+1)
2^(x+2) / 2^(2^4) = 2^(2x -2) / 2^(2x+1)
2^(x+2 - 4) = 2^(2x-2 - 2x -1)
2^(x-2) = 2^(-3)
x-2 = -3
x = -1
2007-07-31 01:30:00 · answer #3 · answered by CPUcate 6 · 0⤊ 1⤋
2^(x+2-4x) = 2^(2x-2-2x-1)
-3x+2 = -3
x = 5/3
---------
Reason: If you have the same base, then you only need to add or subtract exponents.
2007-07-31 01:26:36 · answer #4 · answered by sahsjing 7 · 0⤊ 0⤋