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Question

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


A
5
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B
7
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C
9
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D
10
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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 $$7$$.

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