Question

# The number $$N=6 \log_{10}2+\log_{10}31$$ lies between two successive integers, whose sum is equal to

A
5
B
7
C
9
D
10

Solution

## The correct option is D $$7$$$$N=6\log_{10}2+\log_{10}31$$$$=\log_{10}2^{6}+log_{10}(31)$$$$=\log_{10}(2^{6}.31)$$$$=\log_{10}(64\times31)$$$$=\log_{10}(1984)$$Now $$\log_{10}(10^{3})=\log_{10}(1000)=3$$Similarly$$\log_{10}(10^{4})=4$$Since $$10^{3}<1984<10^{4}$$$$3<\log_{10}(1984)<4$$Hence it lies between $$3$$ and $$4$$.The sum of these two integers is $$7$$.Maths

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