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If `x_(1)` and `x_(2)` are solution of the equation `log_(5)(log_(64)|x|+(25)^(x)-(1)/(2))=2x`, then

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If (x_1, y_1)&(x_2, y_2) are the solutions of the equaltions, (log)_(225)(x)+log_(64)(y)=4a n d(log)_x(225)-(log)_y(64)=1, (log)_(225)x_1dot(log)_(225)x_2=4 b. (log)_(225)x_1+(log)_(225)x_2=6 c. |(log)_(64)y_1-(log)_(64)y_2|=2sqrt(5) d. (log)_(30)(x_1x_2y_1y_2)=12

If (x_1, y_1)&(x_2, y_2) are the solutions of the equaltions, (log)_(225)(x)+log_(64)(y)=4a n d(log)_x(225)-(log)_y(64)=1, (A) (log)_(225)x_1dot(log)_(225)x_2=4 (B). (log)_(225)x_1+(log)_(225)x_2=6 (C). |(log)_(64)y_1-(log)_(64)y_2|=2sqrt(5) (D). (log)_(30)(x_1x_2y_1y_2)=12

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