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The number of digits in 20^(301) (given...

The number of digits in `20^(301)` (given `log_(10)2=0.3010) is

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Let S denotes the antilog of 0.5 to the base 256 and K denotes the number of digits in 6^(10) (given log_(10)2=0.301 , log_(10)3=0.477 ) and G denotes the number of positive integers, which have the characteristic 2, when the base of logarithm is 3. The value of SKG is

Let S denotes the antilog of 0.5 to the base 256 and K denotes the number of digits in 6^(10) (given log_(10)2=0.301 , log_(10)3=0.477 ) and G denotes the number of positive integers, which have the characteristic 2, when the base of logarithm is 3. The value of SKG is

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Let S denotes the antilog of 0.5 to the base 256 and K denotes the number of digits in 6^(10) (given log_(10)2=0.301 , log_(10)3=0.477 ) and G denotes the number of positive integers, which have the characteristic 2, when the base of logarithm is 3. The value of SKG is

Let S denotes the antilog of 0.5 to the base 256 and K denotes the number of digits in 6^(10) (given log_(10)2=0.301 , log_(10)3=0.477 ) and G denotes the number of positive integers, which have the characteristic 2, when the base of logarithm is 3. The value of SKG is

Let S denotes the antilog of 0.5 to the base 256 and K denotes the number of digits in 6^(10) (given log_(10)2=0.301 , log_(10)3=0.477 ) and G denotes the number of positive integers, which have the characteristic 2, when the base of logarithm is 3. The value of SKG is

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Find the number of digit in 4^(2013) , if log_(10)2=0.3010 .