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

Answer 1

Answer: (B)

7

Answer 2

Using the logarithm property $a \log_b c = \log_b c^a$, we can rewrite the expression for $N$ as: $N = \log_{10}(2^6) + \log_{10}31$ $N = \log_{10}64 + \log_{10}31$ Using the logarithm property $\log_b c + \log_b d = \log_b (c \times d)$: $N = \log_{10}(64 \times 31)$ Calculating the product $64 \times 31$: $64 \times 31 = 1984$ So, $N = \log_{10}1984$ We need to find two successive integers, $k$ and $k+1$, such that $k < N < k+1$. This means: $k < \log_{10}1984 < k+1$ This inequality is equivalent to: $10^k < 1984 < 10^{k+1}$ Let's examine powers of 10: $10^1 = 10$ $10^2 = 100$ $10^3 = 1000$ $10^4 = 10000$ We observe that $1000 < 1984 < 10000$. Therefore, $10^3 < 1984 < 10^4$ Comparing this with $10^k < 1984 < 10^{k+1}$, we find that $k=3$. The two successive integers between which $N$ lies are 3 and 4. The question asks for the sum of these two successive integers: $3 + 4 = 7$