(2log6+6log2)/(4log2+log27-log9)এর মান কত?

প্রশ্ন: \( \frac{2\log 6 + 6\log 2}{4\log 2 + \log 27 - \log 9} \) এর মান কত?

সমাধান:

আমরা জানি, \( \log a^b = b \log a \) এবং \( \log (ab) = \log a + \log b \)।

তাহলে, \( 2\log 6 + 6\log 2 = 2(\log (2\times 3)) + \log 2^6 = 2(\log 2 + \log 3) + \log 64 = 2\log 2 + 2\log 3 + \log 64 \)

আবার, \( 4\log 2 + \log 27 - \log 9 = \log 2^4 + \log 3^3 - \log 3^2 = \log 16 + \log 27 - \log 9 \)

এখন, লবের অংশ:

\( 2\log 6 + 6\log 2 = 2\log(2\cdot3) + \log(2^6) = 2(\log 2 + \log 3) + \log 64 = 2\log 2 + 2\log 3 + \log 2^6 = 2\log 2 + 2\log 3 + 6\log 2 = 8\log 2 + 2\log 3 \)

হরের অংশ:

\( 4\log 2 + \log 27 - \log 9 = \log(2^4) + \log(3^3) - \log(3^2) = \log 16 + \log 27 - \log 9 = \log 2^4 + \log 3^3 - \log 3^2 = 4\log 2 + 3\log 3 - 2\log 3 = 4\log 2 + \log 3 \)

অতএব, \( \frac{2\log 6 + 6\log 2}{4\log 2 + \log 27 - \log 9} = \frac{8\log 2 + 2\log 3}{4\log 2 + \log 3} = \frac{2(4\log 2 + \log 3)}{4\log 2 + \log 3} = 2 \)

সুতরাং, উত্তর: 2 🎉