# 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

• @coffeemath I already corrected it Sep 27, 2020 at 21:01
• I saw that (have already erased comment). Sep 27, 2020 at 21:04
• @coffeemath Could you tell me how to do it? Sep 27, 2020 at 21:05
• See my answer... let me know if more explanation needed. Sep 27, 2020 at 21:11

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.

• Could be like this? $x\in \bigcap _{ j=1 }^{ \infty }{ ({ -10 }^{ -10j },{ 10 }^{ -10j }) }$ Sep 27, 2020 at 21:14
• Yes, then finish by actually computing that intersection. Sep 27, 2020 at 21:30

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.

• I see, it's a very interesting suggestion, thank you very much. Sep 27, 2020 at 21:22
• Can you give me a suggestion for this problem? $3^{-j}\leq x \leq 2^{-j}$ Sep 27, 2020 at 21:41
• This one has no specific answer for all the j's because you can be tempted to choose $x = 0$ again by the same reasons than the previous one. However $\frac{1}{3}\leq 0 \leq \frac{1}{2}$ is false. And if you choose something bigger than 0, the limit when $j$ goes to $\infty$ is the following one: $0\leq x\leq 0$ which is false again. Sep 27, 2020 at 21:47
• So how could it end? Sep 27, 2020 at 21:48
• Could you conclude by saying: not true for any value of x? Sep 27, 2020 at 21:49

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

$$10^{-\frac{10}{j}}\le x \le 1$$

$$10^{-5}=0.00001\quad\land\quad 0.00001^2=10000000000=10^{10}$$ $$10^{-2}=0.01\quad\land\quad 0.01^5=10000000000=10^{10}$$ $$10^{-1}=0.1\quad\land\quad 0.1^{10}=10000000000=10^{10}$$