How many positive integers $n$ are there such that $10^n \leq n^{10}$ ?

The answer is 9.

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Only 2 correct solutions were submitted, of which this is one of them. Students had major difficulties justifying their opinions, and even wrote contradictory statements.

The most common argument was that $10^n$ 'obviously' grows faster than $n^{10}$ . If this statement is true, it would contradict the fact that $n=1$ is not a solution but $n=2$ is a solution, which they were all too happy to mention in the next paragraph. Students who even tried to justify this "by differentiating" clearly didn't do the differentiation, or did it wrong (as in several cases). The statement that "Exponential growth is faster than Polynomial growth" is only true for sufficiently large $n$ , and not necessarily true for small values.

Another common mistake made was in not understanding the graph $\frac {n}{\log n}$ completely, and attempting to calculate when this is $>10$ . It can be difficult to justify that $\frac {n}{\log n} = 10$ only has 2 roots, unless you really understand the behavior of this graph.

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So what's the correct justification and solution to this?

Shanu Jindal
- 3 years, 3 months ago

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Due to the fact that the logarithm function is strictly increasing then we can change the inequality to

$n \le 10\, \log_{10}n.$

Those two functions take the same value for $n=\{1.37....,10\}$ . Although one can not know immediately the first solution at least one can deduce it is between 1 and 2.

Solutions need to be between 2 and 10, because values of $n$ greater than 10 do not satisfy the proposed condition.

So the final answer is 9, corresponding to the positive integers between 2 and 10, both included.

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Transform the inequality by taking the log base 10 of both sides to see $n \leq 10 \log n$ . We can do this because log is defined on all positives, is one-to-one, and is increasing. Note, we have equality when $n$ is 10. Furthermore, $n$ grows faster than $\log n$ so 10 must be the largest $n$ for which the inequality holds.

We just need to check for $n$ less than 10, and we readily see that only when $n$ is 1, does the inequality not hold.

So the solution set is all positive integers less than or equal to 10 except. Then counting on our fingers we see that there are 9 such integers.

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The first important thing to note is that the smallest such value for n is 2, since 10^1>1^10. Next, note that since the equation asks for a value greater than OR EQUAL to 10^n. So, it is easy to see both sides ate equal when n=10, since both sides are 10^10. Finally, we note that 10^n increases significantly faster than n^10. This can be shown by taking their derivatives:

(d/dn)10^n at n=10 is ln10
*
10^10.
(d/dn)n^10 at n=10 is 10
*
10^9= 10^10.

Since ln10>1, 10^n>n^10 if n>10. So, the solutions are the integers 2-10 inclusive, which consist of 9 numbers.

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We can check that the integers 2 to 10 satisfy the condition. For $n \geq 11$ , we will show by induction that $10^n > n^{10}$ . Going from $k$ to $k+1$ , we are multiplying the LHS by 10, and multiplying the RHS by $\left(\frac {k+1}{k}\right)^{10}$ , which is less than or equal to $\left(\frac {11}{10} \right)^{10}$ , which is less than 10. (For example, $1.1^5 = 1.61051 < 3$ , implying $1.1^{10} < 9$ .)

Hence, there are $10-2+1 = 9$ such positive integers.

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TenToN = function(n){ for(var i=0; i<n; i++){ var temp=Math.pow(10, i); var temp2=Math.pow(i, 10); //console.log(i+" "+temp+" "+temp2);

```
if(temp<=temp2){
console.log(i+" "+temp+" "+temp2);
}
}
```

}

TenToN(1000);

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```
10^n n^10 10^n-n^10
```

1 10- 1= 9 >0
2 100- 1024= -924 <0
3 1000 -59049= -58049 <0
4 10000 -1048576 =-1038576<0
5 100000 -9765625= -9665625 <0
6 1000000 -60466176 =-59466176<0
7 10000000 -282475249 =-272475249 <0
8 100000000 -1073741824 =-973741824 <0
9 1000000000 -3486784401 =-2486784401<0

10 10000000000 -10000000000= 0

11 1E+11 -25937424601 =74062575399 >0
12 1E+12 61917364224 9.38083E+11
13 1E+13 1.37858E+11 9.86214E+12
14 1E+14 1E+14

15 1E+15 1E+15

16 1E+16 1E+16

17 1E+17 1E+17

18 1E+18 1E+18

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Since we know that 10^10 = 10^10, the maximum value of n must be 10. Also, we know that n cannot be 1 because 1 is not greater that 10. Hence the smallest possible value of n will be 2, because 1024 is more that 100.

Thus, n can be any number from 2 to 10, or 10-2 + 1 = 9 numbers.

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The answer is 9, for values $2,3,..., 10$ . Note that that $10^n > n^{10}$ for $n = 1, 11$ and $10^n \geq n^{10}$ for $2\leq n\leq 10$ . We claim that $10^n > n^{10}$ for $n > 11$ . This can be shown by induction, since we have $n = 11$ as a base case. If $10^n > n^{10}$ and $n \geq 11$ then $10^{n+1} = 10\cdot 10^n > 10n^{10} = (n\sqrt[10]{10})^{10} > (n+1)^{10}$