For what values ​of x is true ${-(10^{10}})^{-j}\leq x \leq ({10^{10}})^{-j}$ whit $j=1,2,3...$ Can you give me any suggestions?
I understand that it is the same as
$x\leq|({10^{10}})^{-j}|$
but I don't know how to conclude
 A: It depends on how you interpret the final phrase "with $j=1,2,3,\cdots.$" If it means just pick one of them then your answer is OK. If it means it should be true for every $k$ you need to intersect all the answers.
A: The bigger the j is, the smaller your x should be. Therefore, if the question is for which values of x, $x≤|(10^{10})^{−j}|$ is always true no matter the j you're choosing, then the answer will be $0\leq x\leq 0$. You can prove this answer is correct by showing the limit of $|(10^{10})^{−j}|$ when j goes to $\infty$ is a number closer and closer to 0.
A: If we extend the examples
$$1^{-1}=1\qquad0.1^{-1}=10\qquad 0.01^{-1}=100\qquad 
0.001^{-1}=100\qquad 0.0001^{-1}=10000\qquad $$
we can see that
$${10}^{-10}\le x\le 1$$
if and only if $j=1.\quad$
For $j=2\implies 10^{-5}\le x \le 1$ and the trend is not exact with j-values that do not divide 10 evenly bu it works out if $j|10$, then
$$10^{-\frac{10}{j}}\le x \le 1$$
$$10^{-5}=0.00001\quad\land\quad 0.00001^2=10000000000=10^{10}$$
$$10^{-2}=0.01\quad\land\quad 0.01^5=10000000000=10^{10}$$
$$10^{-1}=0.1\quad\land\quad 0.1^{10}=10000000000=10^{10}$$
