Without using tables or a calculator, solve the equation 4^x + 2^ 2(x - 1) = 160
How to do? I don't even understand...
2007-10-02 19:12:32 · 5 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
4^x + 2^(2x-2) = 160
the 4 can be changed to 2^2 and get:
2^2x + 2^(2x-2) = 160
now break up the second term:
2^2x + (1/4)2^2x = 160
factor
2^2x (1+1/4) = 160
2^2x (5/4) = 160
2^2x = 128
2^2x = 2^7
2x=7
x=3.5
2007-10-02 19:31:56 · answer #1 · answered by 037 G 6 · 0⤊ 0⤋
4^x + 2^(2(x - 1)) = 160
4^x + (2^2)^(x - 1) = 160
4^x + 4^(x - 1) = 4^2 * 10
(4^x + 4^(x - 1))/(4^2) = 10
4^(x - 2) + 4^(x - 1 - 2) = 10
4^(x - 2) + 4^(x - 3) = 10
4^(x - 2) + 4^(x - 3) = 5 * 2
(4^(x - 2) + 4^(x - 3))/2 = 5
((2^2)^(x - 2) + (2^2)^(x - 3))/2 = 5
(2^(2x - 4) + 2^(2x - 6))/(2^1) = 5
2^(2x - 4 - 1) + 2^(2x - 6 - 1) = 5
2^(2x - 5) + 2^(2x - 7) = 5
(2^(2x - 5))(1 + 2^(-2)) = 5
2^(2x - 5) * (1 + (1/4)) = 5
2^(2x - 5) * (5/4) = 5
2^(2x - 5) = 4
2^(2x - 5) = 2^2
2x - 5 = 2
2x = 7
x = (7/2)
Sorry i wrote this so long, i just wanted to show you little by little.
ANS : x = (7/2)
2007-10-02 19:56:11 · answer #2 · answered by Sherman81 6 · 0⤊ 0⤋
This is sneaky because you have to know your exponents and how to "play" with them. Because 2^2=4 and 2^-2= 1/4, we can rewrite this as 2^2x + [2^2x]/4 =160 or 5/4 * 2^2x= 160.
Then 2^2x= 128. Now 128 is 2 to the 7th power.
So then 2x=7 and x=3.5.
2007-10-02 19:29:44 · answer #3 · answered by cattbarf 7 · 0⤊ 1⤋
well all I can ad is
solve for the brackets first../
2007-10-02 19:21:34 · answer #4 · answered by Anonymous · 0⤊ 1⤋
try to do that step by step
2007-10-02 19:18:31 · answer #5 · answered by tdapower 3 · 0⤊ 1⤋