Solve (3*(-1/3)^(x-1))^2<0.001 ?

1 Answer
Oct 30, 2017

x ge 7 an odd integer

Explanation:

To have a real solution we must have x-1={1,2, 4, cdots, 2k}

For x-1=1 we have

(3 * (-1/3))^2 = 1 > 0.001

For x-1 = 2 we have

(3*(-1/3)^2)^2 = 1/9 > 0.001

and for x-1 = 2k

(3*(-1/3)^(2k))^2 = (3*1/9^k)^2 = 9* 1/9^(2k) = 1/9^(2k-1) < 0.001 or

9^(2k-1) > 1/0.001 = 10^3 and now applying log_10 to both sides

(2k-1)log_10 9 gt 3 or k > 1/2(3/log_10 9+1) approx 2.07193 but k must be integer then

k = 3 and finally

x-1=2*3 rArr x ge 7