I'm new to this website. I've checked the other topics and couldn't find the answer to my question. I've also tried to find the result with TI-CAS but the only thing it does is simplify the inequality which really doesn't help at all.

the equation is : $1000n^3 < 2^n$ Now I've been trying to use log but it gets me nowhere (I haven't seen this form before), the answer is supposed to be 24 but I'm not sure. So my question is how to solve step by step this inequality for n. Could you guys help me with this one?

  • $\begingroup$ your inequality can not be solved for $n$ explicitely you will need a numerical method $\endgroup$ – Dr. Sonnhard Graubner Nov 23 '17 at 20:17
  • $\begingroup$ What does it mean? $\endgroup$ – Yoni Nov 23 '17 at 20:20
  • $\begingroup$ Your problem is not very clearly stated. No, your problem is not stated at all, but one can puzzle out what you are asking. You might have asked "What is the smallest $n$ such that $1000n^2<2^n$, and how do you find it?", in the body of your post, which would have been shorter and clearer than what you wrote. Which would mean we could help you quicker. $\endgroup$ – kimchi lover Nov 23 '17 at 20:32
  • $\begingroup$ ok sorry for that, yes indeed, the question is more for what value of n this inequality is true and how to find it? $\endgroup$ – Yoni Nov 23 '17 at 20:36
  • $\begingroup$ What Dr. Graubner is saying is that we can't give you a nice "step-by-step" method that will just give you the answer. The truth is, there are very few problems where there is such a method that gives a definite answer. In the vast majority of problems, it just isn't possible. Instead you use numerical methods, which only give approximate answers, but where the approximation can (usually) be improved to whatever accuracy you desire and are willing to do the work to accomplish. $\endgroup$ – Paul Sinclair Nov 24 '17 at 0:42

As noted in the comments, there no way to just solve for $n$. So we have to do some investigation of the problem.

Let $f(x) = 2^x - 1000x^3$, then $$f'(x) = 2^x\ln 2 - 3000x^2\\ f''(x) = 2^x(\ln 2)^2 - 6000x\\ f'''(x) = 2^x(\ln 2)^3 - 6000$$

  • $f'''(x_m) = 0$ is solvable to find $x_m = \log_2 6000 - 3\log_2(\ln 2) \approx 14.137$. For lower $x, f'''(x) < 0$, and for higher $x, f'''(x) > 0$.
  • So $f''(x)$ is decreasing to the left of $x_m$ and increasing to the right. Now $f''(0) = (\ln 2)^2 > 0, f''(x_m) \approx -76166 < 0, f''(18) \approx 17948 > 0$, so $f''$ has two zeros $x_l, x_u$ with $0 < x_l < x_m < x_u < 18$. Below $x_l$, $f''$ is positive. From $x_l$ to $x_u$, it is negative, and above $x_u$, it is positive again.
  • So $f'(x)$ is increasing below $x_l$, decreasing from $x_l$ to $x_u$, and increasing above $x_u$. $f'(0) = \ln 2 > 0$, and since $f'$ is increasing here, we can know that $f'(x_l) > 0$. But $f'(1) 2\ln(2) - 3000 < 0$. You can also test that $f'(20) < 0$ and $f'(21) > 0$. Therefore $f'(x) = 0$ has two solutions that are greater than $0$, call them $x_a, x_a$, with $0 < x_a < 1$ and $20 < x_b < 21$. Below $x_a, f'(x) > 0$. Between $x_a$ and $x_b, f'(x) < 0$, and above $x_b, f'(x) > 0$.
  • Finally we get to $f(x)$. It is increasing below $x_a$, decreasing from $x_a$ to $x_b$, and increasing above $x_b$. Since $f(0) > 0$ and $f(1) < 0$, there is one root between $0$ and $1$. $f(x)$ is then decreasing all the way to $x_b$, where it is therefore still below $0$. After $x_b$, it increases, so it can only have one more $0$.

What did all that give us? If we restrict $n \ge 1$, there is only one crossing that needs found, and further, it must be $> 20$. How can you find it? The quickest easy way is to use Newton's method:

  • Start with an initial guess $x_0$.
  • Repeatedly calculate $$x_{m+1} = x_m - \frac{f(x_m)}{f'(x_m)}$$

In this case, $$x_{m+1} = x_m - \frac{2^x - 1000x^3}{2^x\ln 2 - 3000x^2}$$

However, if one starts at $x_0 = 20$, it actually finds the root between $0$ and $1$ instead of the one greater than $20$. $x_0 = 21$ jumps off into high values before heading back down. $x_0 = 22$ gives a reasonable approach: $$\begin{array} {c|c}\text x_m & f(x_m)\\\hline 22 &-6453696\\ 26.43470699 & 72234368.06\\ 25.24618752 & 23706697.72\\ 24.32280169 & 6594885.345\\ 23.80638039 & 1178033.813\\ 23.66727007 & 64678.86294\\ 23.65870725 & 228.9771537\\ 23.65867672 & 0\end{array}$$ (to the limits of Excel's ability to track it.) Thus $n = 24$ is indeed the lowest integer greater than $0$ for which $2^n > 1000n^3$.


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