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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`

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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)=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

Comprehension (Q. No. 6 to Q. No. 8) Let (x_1, y_1)&(x_2, y_2) are the solutions of the equation, (log)_(225)(x)+(log)_(64)(y)=4 and (log)_x(225)-(log)_y(64)=1 (log)_(225)x_1dot(log)_(225)x_2= 2 (b) 4 (c) 6 (d) 8

If (x_(1),y_(1)) and (x_(2),y_(2)) are the solution of the system of equation log_(225)(x)+log_(64)(y)=4 and log_(x)(225)-log_(y)(64)=1, then show that the value of log_(30)(x_(1)y_(1)x_(2)y_(2))=12

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