log_(10)2=0.3010

Given log_(10)2=0.3010,log_(10)3=0.4771 . List-II represents number of significant digits before decimal of entries in List-I. |{:(,"List-I",,"List-II"),((P),5^(12)xx3^(4),(1),4),((Q),2^(10)xx3^(5),(2),6),((R ),(2^(15))/(3^(2)),(3),7),((S),(9^(8))/(2^(3)),(4),11):}|

CH_(4) is adsorbed on 1g charcoal at 0^(@)C following the Freundlich adsorption isotherm. 10.0 mL of CH_(4) is adsorbed at 100 mm of Hg, whereas 15.0 mL is adsorbed at 200 mm of Hg. The volume of CH_(4) adsorbed at 300 mm of Hg is 10^(x) mL . The value of x is _______ xx 10^(-2) . (Nearest integer) [Use log_(10)2 = 0.3010, log_(10)3 = 0.4771 ]

Given that log_10 2=.3010 , log_10 3=.4771 ,then find the value of log_10 ((24)^(1/2))

If log 2 = 0.3010, then log 5 = ______.

If log 2= 0.3010 , and log 3= 0.4771 then log 150 = ______

If log2=0.3010 then the Arithmetic mean of log40 and log 5 is