JEE Advance - Mathematics (2018 - Paper 1 Offline - No. 7)
The value of $${({({\log _2}9)^2})^{{1 \over {{{\log }_2}({{\log }_2}9)}}}} \times {(\sqrt 7 )^{{1 \over {{{\log }_4}7}}}}$$ is ....................
Answer
8
Explanation
$${({({\log _2}9)^2})^{{1 \over {{{\log }_2}({{\log }_2}9)}}}} \times {(\sqrt 7 )^{{1 \over {{{\log }_4}7}}}}$$
$$ = {({\log _2}9)^{{{2.{{\log }_2}} \over {{{\log }_2}9}}}} \times {7^{{1 \over 2}.{{\log }_7}4}}$$
$$ = {({\log _2}9)^{{{\log }_{\log {2^{{9^{{2^2}}}}}}}}} \times {7^{{{\log }_7}2}}$$
$$ = {2^2} \times 2 = 8$$
$$ = {({\log _2}9)^{{{2.{{\log }_2}} \over {{{\log }_2}9}}}} \times {7^{{1 \over 2}.{{\log }_7}4}}$$
$$ = {({\log _2}9)^{{{\log }_{\log {2^{{9^{{2^2}}}}}}}}} \times {7^{{{\log }_7}2}}$$
$$ = {2^2} \times 2 = 8$$
Comments (0)
