Solve ${(3 \cdot {(- \frac{1}{3})}^{x - 1})}^{2} < 0.001$ $(3*(-1/3)^(x-1))^2<0.001$ ?

Question

Solve ${(3 \cdot {(- \frac{1}{3})}^{x - 1})}^{2} < 0.001$ $(3*(-1/3)^(x-1))^2<0.001$ ?

1 Answer

Impact of this question

1366 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-10-30T11:52:41

$x \geq 7$ $x ge 7$ an odd integer

Explanation:

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

For $x - 1 = 1$ $x-1=1$ we have

${(3 \cdot (- \frac{1}{3}))}^{2} = 1 > 0.001$ $(3 * (-1/3))^2 = 1 > 0.001$

For $x - 1 = 2$ $x-1 = 2$ we have

${(3 \cdot {(- \frac{1}{3})}^{2})}^{2} = \frac{1}{9} > 0.001$ $(3*(-1/3)^2)^2 = 1/9 > 0.001$

and for $x - 1 = 2 k$ $x-1 = 2k$

${(3 \cdot {(- \frac{1}{3})}^{2 k})}^{2} = {(3 \cdot \frac{1}{9^{k}})}^{2} = 9 \cdot \frac{1}{9^{2 k}} = \frac{1}{9^{2 k - 1}} < 0.001$ $(3*(-1/3)^(2k))^2 = (3*1/9^k)^2 = 9* 1/9^(2k) = 1/9^(2k-1) < 0.001$ or

$9^{2 k - 1} > \frac{1}{0.001} = 10^{3}$ $9^(2k-1) > 1/0.001 = 10^3$ and now applying ${log}_{10}$ $log_10$ to both sides

$(2 k - 1) {log}_{10} 9 > 3$ $(2k-1)log_10 9 gt 3$ or $k > \frac{1}{2} (\frac{3}{{log}_{10} 9} + 1) \approx 2.07193$ $k > 1/2(3/log_10 9+1) approx 2.07193$ but $k$ $k$ must be integer then

$k = 3$ $k = 3$ and finally

$x - 1 = 2 \cdot 3 \Rightarrow x \geq 7$ $x-1=2*3 rArr x ge 7$

Science

Math

Humanities

... and beyond

Solve ${(3 \cdot {(- \frac{1}{3})}^{x - 1})}^{2} < 0.001$ $(3*(-1/3)^(x-1))^2<0.001$ ?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

Solve (3⋅(−13)x−1)2<0.001(3*(-1/3)^(x-1))^2<0.001 ?

1 Answer

Explanation:

Related questions

Impact of this question

Solve ${(3 \cdot {(- \frac{1}{3})}^{x - 1})}^{2} < 0.001$ $(3*(-1/3)^(x-1))^2<0.001$ ?