10^N Less Than N^10

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 $10^{n+1} = 10\cdot 10^n > 10n^{10} = (n\sqrt[10]{10})^{10} > (n+1)^{10}$

For n>10 LHS>RHS We can use induction to prove it. For n=1 we have 10>1 For 2=<n=<10 the inequality is true

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.

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.

Taking the base 10 logarithm of both sides, we get $n\le\log n^{10}$ , or $n\le10\log n$ . We can divide so that the $n$ is only on one side of the equation to get $\frac{n}{\log n}\le10$ . The numerator of this fraction will increase faster than the denominator, so if we find that for some $k$ that the fraction is equal to 10, then for all integers greater than $k$ , the fraction will be greater than 10. This happens at $k=10$ . Thus, we have our upper bound. The lower bound, $2$ , can be easily found by testing the original inequality with the first few values of $n$ . $\frac{n}{\log n}$ will continuously increase from 2 to 10, so all values in between will work as well. The answer is the number of integers between 2 and 10 inclusive, or 9.

step 1: check for 1: 10^1 ? 1^10 ----> 10 >1 so, 1 doesn't belong to the answer. step 2: check by putting n=2, we find that 10^2<2^{10}. obviously, all n<10 belong to the set of our answer. step 3: now, 10^{10}=10^{10}. so, 10 also belongs to the answer. (if you want, you can check by putting n>10 ----> n=11) step 4: neglecting 1, we find this pattern: 10^(n<10) < (n<10)^{10} , 10^(n=10) = (n=10)^{10} , 10^(n>10) > (n>10)^{10} step 5: count the number of elements of the set of answers, i.e., 2,,3,4.......10 which is equal to 10-1=9 step 6: the answer is 9.

For n=1, (10^1)/(1^10)=10/1=10>1, not valid For n=10, 10^10=10^10, equality valid Now, for n=2, (10^2)/(2^10)=(5^2)/(2^8) is proper fraction<1,valid Likely for n=3,4,5,6,7,8,9 it is correct. Now, for n=11, (10^11)/(11^10)=10(10^10)/(11^10) is improper fraction>1, not valid Just like, n=12,13 or more the given relation becomes invalid. Therefore, number of positive integers is 9.

to solve this problem, simple: 1. When we substitute n integer=1, the result will be 10<=1 and surely that's not fulfill the condition of "<=". 2. Start from n=2 until n=9, the result will fulfill the requirement of "< " than .... 3. Next we get n=10, where the result will be equally same 10^10 = 10^10 and still fulfill the requirement of "=" 4.Start from n=11 until n= inf the result has not fulfilled the requirement which ">" than ... So the array of n's integer is = {2,3,4,...,10) which the length is 9.

Taking natural logarithm on both sides, we have: lg 10^n <= lg n^10 n<= 10 lg n n/(lg n) <= 10 Since if n =10, n/(lg n) = 10, then n <= 10. But n = 1 would mean lg n = 0, rendering n/(lg n) undefined, hence there are only 9 possible positive integer values for n (2 to 10 inclusive).

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.

n=1 is obviously not possible Case of n=2, 10^2 < 2^10 Case of n=10, 10^10 = 10^10 that means it holds for 2≤n≤10 until now. Case of n=11 10^11= (10^5 x 10^(1/2))^2 Since the square root of 10 is 3.16227766..., we got 10^11 > 300000^2 > 161051^2 = (11^5)^2 = 11^10, n=11 is false, for n > 11, the LHS will be multiplied by 10 and the RHS will be multiplied by k^10 (where k = (n+1)/n for n >11) and since the k is indirectly propotional to n, the max value of k is 1.1^10 = 2. ... < 10, for n >11 , the inequality will always fail. QED

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.

1^10 = 1, which means 10^1 < 1 is false. On the other hand, if you calculate 10^2 and 2^10 (100 and 1024) you will see that this is right. Same for 3,4,5,...,8,9 and 10! 10^10 = 10^10 so this is right too! Then 10^11 is obviously bigger than 11^10 => The soultions are following: {2,3,4,5,6,7,8,9,10} //that's nine numbers.

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);

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

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.