JEE Advanced 2018 — Paper 1 — Question 33

Let $a = \log_2 9$. Then the first term is $(a^2)^{\frac{1}{\log_2 a}} = a^{\frac{2}{\log_2 a}}$. Using $\frac{1}{\log_2 a} = \log_a 2$, we have $a^{2 \log_a 2} = (a^{\log_a 2})^2 = 2^2 = 4$. For the second term: $(\sqrt{7})^{\frac{1}{\log_4 7}} = 7^{\frac{1}{2} \cdot \frac{1}{\log_4 7}}$. Since $\frac{1}{\log_4 7} = \log_7 4$, the exponent becomes $\frac{1}{2} \log_7 4 = \log_7 4^{1/2} = \log_7 2$. Thus the second term is $7^{\log_7 2} = 2$. Multiplying gives $4 \times 2 = 8$. Therefore, the value of the expression is 8.