The number `N=6^(log_(10)40). 5^(log_(10)36)` is a natural number ,Then sum of digits of N is :

The value of 6^(log_(10)40)*5^(log_(10)36) is

The value of 6^(log_10 40)*5^(log_10 36) is

The value of 6^(log_10 40)*5^(log_10 36) is

Prove that log_n(n+1)>log_(n+1)(n+2) for any natural number n > 1.

Prove that log_n(n+1)>log_(n+1)(n+2) for any natural number n > 1.

Prove that log_n(n+1)>log_(n+1)(n+2) for any natural number n > 1.

Prove that log_(n)(n+1)>log_(n+1)(n+2) for any natural number n>1

The number of N=6-(6(log)_(10)2+(log)_(10)31) lies between two successive integers whose sum is equal to (a)5(b) 7(c)9(c)10

The number of N=6-(6(log)_(10)2+(log)_(10)31) lies between two successive integers whose sum is equal to (a)5 (b) 7 (c) 9 (c) 10

The number of N=6-(6(log)_(10)2+(log)_(10)31) lies between two successive integers whose sum is equal to (a)5 (b) 7 (c) 9 (c) 10