The value of `log_2`16 is
If log(0.57) = `overline(1)`.756, then the value of log 57 + log `(0.57)^3 + log `sqrt(0.57)` is
If log 2 = 0.30103, then the number of digit in `5^20` is :
If log 2 = 0.30103, the number of digit in `4^50` is :
If log 2 = 0.30103, the number of digits in `2^64` is :