Exponential Equations:
(1) If 2^x = 3, find 4^(-x)
(2) If 5^(-x) = 3, find 5^(3x)
2007-12-26 10:20:16 · 4 answers · asked by journey 1 in Science & Mathematics ➔ Mathematics
(1) If 2^x = 3, find 4^(-x)
4^(-x) = (2^2)^(-x) = 2^(-2x) = (2^x)^(-2) = 3^(-2)
1/3^2 = 1/9 ANS
(2) If 5^(-x) = 3, find 5^(3x)
5^(3x) = 5^(-3*-x) = [5^(-x)]^(-3) = 3^(-3) = 1/3^3 = 1/27 ANS
teddy boy
2007-12-26 10:59:07 · answer #1 · answered by teddy boy 6 · 0⤊ 0⤋
(#1) 4^-x = 1/4^x = 1/(2^2)^x = 1/(2^x)^2 = 1/3^2 = 1/9
(#2) 5^(3x) = 1/5(-3x) = 1/5(-x)^3 = 1/3^3 = 1/27
that's it! :)
2007-12-26 18:26:55 · answer #2 · answered by Marley K 7 · 0⤊ 0⤋
1. 2^x = 3 then 4^(--x) = {(2^2)}^(--x) = (2^x)^(--2) = 3^(--2) = 1/3^2 = 1/9
2. 5^(--x) = 3 then 5^(3x) = {5^(--x)}^(--3) = 3^(--3) = 1/3^3 = 1/27
2007-12-26 18:40:39 · answer #3 · answered by sv 7 · 0⤊ 0⤋
(#1) We know that 2^x = 3
Solve for x, so we get xlog2= log3
x=log3/log2
4^(-(log3/log2))= 1/9 [0.11]
(#2) We know that 5^-x=3
Solve for x,, so we get -xlog5=log3
x=-(log3/log5)
5^(3*(-log3/log5))= 1/27 [~0.037]
2007-12-26 18:27:15 · answer #4 · answered by ¿ /\/ 馬 ? 7 · 0⤊ 0⤋