Solve each equation for x.
(1) f(x) = (1/2)^(1 - x) = 4
(2) f(x) = 4^x - 2^x = 0
2007-12-26 10:15:27 · 5 answers · asked by journey 1 in Science & Mathematics ➔ Mathematics
(#1) 0.5^(1-x)=4^1
Change both bases (i.e.: 0.5 and 4) to base 2, so we have
2^-(1-x) = 2^(2)1
2^-1+x = 2^2
Base 2s cancel out so we get
-1+x = 2
x=3
(#2) 4^x - 2^x = 0
rearrange the equation to form 4^x = 2^x
Change both bases to base 2 again so we get
4^x = 2^x
2^(2)x = 2^x
base 2s cancel out so we get
2x = x
2x-x=0
x(2-1)=0
x=0
2007-12-26 10:23:26 · answer #1 · answered by ¿ /\/ 馬 ? 7 · 0⤊ 0⤋
(1) x=3 because 1/2 is 2^-1 so it becomes 2^(x-1) and 4=2^2 so (x-1)=2 and therefore x=3
(2) x=0 only when x=0 does 4^x and 2^x equal the same thing, which is one by t he way
2007-12-26 10:23:27 · answer #2 · answered by Engr Dude 3 · 0⤊ 0⤋
1) You can almost do this by inspection, you know that 1/2 is 2^(-1), and 2^2 =4, so you should expect that the exponent 1-x will equal -2, or that x=3.
To prove this:
take logs of both sides:
(1-x)log2=log4
1-x=1.3863/-0.6931=-2
x=3
2)4^x-2^x=0
4=2^2, so
2^(2x)=2^x => 2x=x which has no solution
please check for a typo, I dont think this has a solution (or maybe my middle aged eyes are not seeing a term...)
2007-12-26 10:22:14 · answer #3 · answered by kuiperbelt2003 7 · 0⤊ 1⤋
I don't think the first one is solvable.
The second one is 0.
2007-12-26 10:19:41 · answer #4 · answered by Anonymous · 0⤊ 1⤋
do ur hw. i did. and why r u doing hw now. its winter vacation.
2007-12-26 10:18:03 · answer #5 · answered by Anonymous · 0⤊ 1⤋