❓ Question Find the value of 81 1 log 5 3 + 27 log 9 36 + 3 4 log 7 9 . 81^{\frac{1}{\log_5 3}} +27^{\log_9 36} +3^{\frac{4}{\log_7 9}}. ✍️ Solution First term 81 1 log 5 3 = ( 3 4 ) 1 log 5 3 = 3 4 log 5 3 . 81^{\frac{1}{\log_5 3}} =(3^4)^{\frac{1}{\log_5 3}} =3^{\frac{4}{\log_5 3}}. Using 1 log b a = log a b , \frac{1}{\log_b a}=\log_a b, we get 3 4 log 3 5 . 3^{4\log_3 5}. Using a k log a b = b k , a^{k\log_a b}=b^k, 3 4 log 3 5 = 5 4 = 625. 3^{4\log_3 5}=5^4=625. Second term 27 log 9 36 = ( 3 3 ) log 9 36 = 3 3 log 9 36 . 27^{\log_9 36} =(3^3)^{\log_9 36} =3^{3\log_9 36}. Now, 36 = 6 2 , 9 = 3 2 . 36=6^2,\qquad 9=3^2. Hence, 3 log 9 36 = 3 ⋅ log 3 36 log 3 9 = 3 ⋅ 2 log 3 6 2 = 3 log 3 6. 3\log_9 36 = 3\cdot\frac{\log_3 36}{\log_3 9} = 3\cdot\frac{2\log_3 6}{2} = 3\log_3 6. Therefore, 3 3 log 3 6 = 6 3 = 216. 3^{3\log_3 6}=6^3=216. Third term 3 4 log 7 9 = 3 4 log 9 7 .