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

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Answer 1 · 2017-10-30T11:52:41

#x ge 7# an odd integer

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#