Hi guys I have an equation like this: $$2n-n^8<0$$

How can I simplify this equation and get the value of $n$?

  • 1
    $\begingroup$ What do you mean by "the value of $n$"? There are infinitely many values of $n$ that satisfy the inequality $2n-n^8<0$. $\endgroup$ – user5137 Mar 18 '12 at 3:50
  • $\begingroup$ @jackMAney- Yeah my bad..It should be the highest vale of n for which the condition is satisfied $\endgroup$ – karthick Mar 18 '12 at 4:26

$2n-n^8\lt 0$ is equivalent to $n^8 - 2n \gt 0$. Since $n^8 -2n = n(n^7-2)$, you are looking at $$ n(n^7-2)\gt 0.$$

A product is positive if both factors are positive, or if both factors are negative. So:

Case 1. Both factors positive.

We need $n\gt 0$ and $n^7-2\gt 0$; $n^7-2\gt 0$ means $n^7\gt 2$; taking $7$th roots (which we can do because $7$ is odd, so raising to the $7$th power respects inequalities) we get $n\gt 2^{1/7}$. So we need $n\gt 0$ and $n\gt 2^{1/7}$. The latter condition implies the first, so we get that this case will happen exactly when $n\gt 2^{1/7}$.

Case 2. Both factors negative.

I'll let you do this one.


First of all what you have is an inequality, not an equation, as you claim.

So, $$\begin{align}2n-n^8 &\lt 0 \\ n^8 & \gt2n \end{align}$$

To determine all the solutions to this inequality, we go over every value of $n$ and ask if this $n$ satisfies our inequality. We break this process into various cases: $n \gt 0$; $n \lt 0$ and $n=0$

Case 1: $n \lt 0$

Note that if $n \lt 0$, we have that $2n \lt 0$ but however, $n^8 \gt 0$. This means, $$n^8 \gt 0 \gt 2n ~~~~\mbox{for all $n \lt 0$}$$

Case 2: $n \gt 0$

If $n \gt 0$, we can cancel out the $n$ without having to reverse the inequality. So, we get, $$n^7 \gt 2 \implies n \gt \sqrt[7] 2$$

Case 3: $n=0$

It is also clear that $n=0$ is not a solution as $0 \not \gt 0$.

So, the set of all $n$ that satisfies this inequality is $$n \in (-\infty, 0) \cup (\sqrt[7]{2}, \infty)$$


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