14092=0.30103" then "solve2^(x)=25

If log 2=0.30103 then solve 2^(x)=25

If log2=0.30103 , then log32=….

Solve:- 5x = 25

If log_(10)2=0.30103, find the value of log_(10)50.

If log_(10)2=0.30103 find the value of log_(5)32

Solve: x+1/x=25

Ther are n zeros appearing immediately after the decimal point in the value of (0.2)^(25) . It is given that the value of "log"_(10)2=0.30103 . The value of n is

Solve : (2x-3)^(2)=25 .