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The number of `N=6-(6(log)_(10)2+(log)_(10)31)`lies between two successive integers whose sum is equal to(a)5 (b) 7 (c) 9 (c) 10 |
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Answer» `N in(I_1,I_2)` `N=6-[6log_10^2+log_10^31]` `N=6-[log_10^(2^6)+log_10^31]` `N=6-[log_10^32+log_10^31]` `N=6-[log_10^(32*31)]` `N=6-log_10^992` `N=6-log_10^(10^3-8)]` `N=6-(3^-)` `N=6-(3-h)` `N=6-3+h` `N=(3+h)` Sum=3+4=7. |
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