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if 11 x = 3 y = 99 z then find (1/x) + (1/y) + (1/z) = ? 1) (2/z) – (1/y) 2) (2/z) + (1/y) 3) -1/y 4) 0
11x=3y=99zthis implies that, xlog11=ylog3=zlog99(1/x)+(1/y)+(1/z)=(log11/ylog3) + (1/y) +(log99/ylog3) {by putting x=ylog3/log11 and z=log99/ylog3}now taking out 1/ylog3 common it becomes=(1/ylog3)(log11+log3+log99)=log(11*33*99)/ylog3 Now,(2/z)-(1/y)=2log99/xlog11- log3/xlog11 {similarly by putting z=xlog11/log99 and y=xlog11/log3} =log(99*99/3)/xlog11 =log(99*33)/ylog3 =log(99*3*11)/ylog3 =(1/x) +(1/y) +(!/z) therefore the answer should be option(1)please approve if useful ; )
Taking log both sides we get xlog11=ylog3=zlog99We have to find 1/z+1/y+1/x-(1)So....Putting values of x,y,z in (1) we will get answer Option (A) I.e.2/z-1/y
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