(2log6+6log2)/(4log2+log27-log9)= কত?

Explanation:

Another Explanation (5): সমাধান: প্রদত্ত রাশি: \(\frac{2\log 6 + 6\log 2}{4\log 2 + \log 27 - \log 9}\) আমরা জা??ি, \(n \log a = \log a^n\) এবং \(\log a + \log b = \log (ab)\) সুতরাং, লব: \(2\log 6 + 6\log 2 = \log 6^2 + \log 2^6 = \log 36 + \log 64 = \log (36 \times 64) = \log 2304\) হর: \(4\log 2 + \log 27 - \log 9 = \log 2^4 + \log 3^3 - \log 3^2 = \log 16 + \log 27 - \log 9 = \log (16 \times 27) - \log 9 = \log 432 - \log 9 = \log \frac{432}{9} = \log 48\) এখন, \(\frac{2\log 6 + 6\log 2}{4\log 2 + \log 27 - \log 9} = \frac{\log 2304}{\log 48}\) আমরা লিখতে পারি, \(2304 = 48^2\) অতএব, \(\frac{\log 2304}{\log 48} = \frac{\log 48^2}{\log 48} = \frac{2 \log 48}{\log 48} = 2\) সুতরাং, \(\frac{2\log 6 + 6\log 2}{4\log 2 + \log 27 - \log 9} = 2\) 🥳🎉