log2 2a3−b3=61 (where a and b are consecutive natural numbers and a greater than b ) , then the value of a is
3
4
5
7
log22a3−b3 = a3−b3If a and b are consecutive natural numbers then a3−b3 = 1+3ab
61 = 1 + 3 ×4 × 5
So, a = 5
The value of log(a2−2ab+b2), (where a and b are consecutive natural numbers and a is greater than b) is
The number of integers of the form 3AB4, where A, B denote some digits, which are divisible by 11 is
A. 0
B. 4
C. 7
D. 9
If 2 is a zero of both the polynomial, 3x2+ax−14 and 2x−b, then value of (a−2b) is,