Just write it as 2 to the power n. You can't simplify that, and you can only calculate a specific value if you know the value of n.
3 x10 30 x10
3
Oh, dude, 2 to the power of n minus 1 calculates the largest Mersenne prime number less than 2 to the power of n. It's like the cool kid of prime numbers, you know? Just strutting around, being all big and powerful. So yeah, that's what it does.
It is: 2 to the power of 8 = 256
2
The one line expression is: ((0 != n) && !(n & n-1)) example: int main () { for (int n = 0; n <= 1000001; ++n) { if ((0 != n) && !(n & n-1)) cout << n << " is a power of 2" << endl; } return 0; } will produce: 1 is a power of 2 2 is a power of 2 4 is a power of 2 8 is a power of 2 16 is a power of 2 32 is a power of 2 64 is a power of 2 128 is a power of 2 256 is a power of 2 512 is a power of 2 1024 is a power of 2 2048 is a power of 2 4096 is a power of 2 8192 is a power of 2 16384 is a power of 2 32768 is a power of 2 65536 is a power of 2 131072 is a power of 2 262144 is a power of 2 524288 is a power of 2
n is the no. so therefore a n value is the numerical or fix value e.g. 7^n, where n is 2 your result will be 49.
2^(1) = 2 Remember this table of powers. n^(0) = 1 n^(1) = n n^(2) = n^(2) e.g. 2^(0) = 1 2^(1) = 2 2^(2) = 4 et seq., NB When the index number of '1' is used , it is only there as a guide to manipulating indices ( index numbers).
3^n+2+3^n = 6^n+2 *'to the power of' can be represented with this symbol ^ .
puts ("32"); /*or*/ printf ("2^5=%d\n", 1<<5);
The value of 4 to the power of -2 is 0.0625
16(2^n)(10)(2^n)=160[2^(2n)]=160(4^n)
Assume 2^k < k! for all n > k here n > 2, then 2^n = 2^(n - 1)*2 < (n-1)! * n = n! Done. Connie and John
When the equation 2 raised to the power of log n is simplified, it equals n.
n is the no. so therefore a n value is the numerical or fix value e.g. 7^n, where n is 2 your result will be 49.
N+2
2/7 = n/14