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The solutions to the system of equations `log_5 x+log_(27) y=4` and `log_x5-log_y(27)=1` are `(x_1,y_1)` and `(x_2,y_2)` then `log_(15) (x_1x_2y_1y2)` is

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1+log_(x)y=log_(2)y

If (x_(1),y_(1))&(x_(2),y_(2)) are the solutions of the equaltions,log_(225)(x)+log_(64)(y)=4 and log_(x)(225)-log_(y)(64)=1log_(225)x_(1).log_(225)x_(2)=4 b.log_(225)x_(1)+log_(225)x_(2)=6c|log_(64)y_(1)-log_(64)y_(2)|=2sqrt(5)d*log_(30)(x_(1)x_(2)y_(1)y_(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, (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