`log_(3/4)log_8(x^2+7)+log_(1/2)log_(1/4)(x^2+7)^(-1)=-2`.

1.	`log_(3/4)log_8(x^2+7)+log_(1/2)log_(1/4)(x^2+7)^(-1)=-2`.
Answer» Correct Answer - `pm3` `log_(3//4)log_(8)(x^(2)+7)+log_(1//2)log_(1//4)(x^(2)+7)^(-1)=-2` `rArr log_(3//4)((1)/(3)log_(2)(x^(2)+7))-log_(2)(log_(2)(x^(2)+7))/(2)=-2` Let `log_(2)(x^(2)+7)=t` `rArr "log"_(3//4)(t)/(3)-"log"_(2)(t)/(2)+2=0` `rArr "log"_(3//4)(t)/(3)+1-("log"_(2)(t)/(2)-1)=0` `rArr "log"_(3//4)(t)/(4)="log"_(2)(t)/(4)` `rArr (t)/(4)=1` `rArr t = 4` `rArr log_(2)(x^(2)+7)=4` `rArr x^(2)+7=16` `rArr x^(2)=9` `rArr x^(2)= pm 3`

Discussion

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