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In a binomial distribution B(n , p=1/4) ...

In a binomial distribution `B(n , p=1/4)` , if the probability of at least one success is greater than or equal to `9/(10)` , then n is greater than (1) `1/((log)_(10)^4-(log)_(10)^3)` (2) `1/((log)_(10)^4+(log)_(10)^3)` (3) `9/((log)_(10)^4-(log)_(10)^3)` (4) `4/((log)_(10)^4-(log)_(10)^3)`

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In a binomial distribution B(n,p=(1)/(4)), if the probability of at least one success is greater than or equal to (9)/(10) ,then n is greater than (1)(1)/(log_(10)4)(2)(1)/(log_(10)^(4)+log_(10)^(3))(3)(9)/(log_(10)^(4)-log_(10)^(3))(4)(4)/(log_(10)^(4)-log_(10)^(3))

(1)/(3)log_(10)125-2log_(10)4+log_(10)32

log4^2_10 + log4_10

((log)_(2)3)(log)_(3)4(log)_(4)5(log)_(n)(n+1)=10 Find n=?

log_(10)15(1+log_(15)30)+(1)/(2)log_(10)16(1+log_(4)7)-log_(10)6(log_(6)3+log_(6)7)

log_(10)((384)/(5))+log_(10)((81)/(32))+3log_(10)((5)/(3))+log_(10)((1)/(9))

log_(10)(log_(2)3) + log_(10)(log_(3)4) + …….. + log_(10) (log_(1023) 1024) equals

What is the value of log_(10)(9/8)-log_(10)((27)/(32))+log_(10)(3/4) ?

What is the value of (1/3 log_(10)125 - 2log_(10)4 + log_(10)32 + log_(10)I) ?

Solve : (iii) 4^(log_(9^3))+9^(log_(2^4))=10^(log_(x^83))