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Question

If logxbc=logyca=logzab then prove that xb+c.yc+a.zb+a=1.

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Solution

Given,
logxbc=logyca=logzab=k(let). [ Where k is variation constant]
Then logx=k(bc)
or, x=ek(bc).....(1) and similarly y=ek(ca)......(2) and z=ek(ab).....(3).
Now,
xb+c.yc+a.zb+a
=ek(b2c2).ek(c2a2).ek(a2b2)
=ek(b2c2+c2a2+a2b2)
=e0
=1.

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