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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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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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Given system of simultaneous equations 2^x . 5^y=1 a n d 5^(x+1). 2^y=2. Then - a.x=(log)_(10)5 a n d y=(log)_(10)2 b. x=(log)_(10)2 a n d y=(log)_(10)5 c. x=(log)_(10)(1/5) a n d y=(log)_(10)2 d. x=(log)_(10)5 a n d y=(log)_(10)(1/2)

Statement-1: The solution set of the equation "log"_(x) 2 xx "log"_(2x) 2 = "log"_(4x) 2 "is" {2^(-sqrt(2)), 2^(sqrt(2))}. Statement-2 : "log"_(b)a = (1)/("log"_(a)b) " and log"_(a) xy = "log"_(a) x + "log"_(a)y

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