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Question

If lnxbc=lnyca=lnzab, then which of the following option(s) is/are true ?

A
xyz=1
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B
xaybzc=1
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C
xb+c yc+a za+b=1
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D
xyz=xaybzc
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Solution

The correct options are
A xyz=1
B xaybzc=1
C xb+c yc+a za+b=1
D xyz=xaybzc
Let lnxbc=lnyca=lnzab=k (say)
lnx=k(bc), lny=k(cb), lnz=k(ab)

lnx+lny+lnz=k(bc)+k(cb)+k(ab)=0
lnxyz=0
xyz=1

Also, x=ek(bc), y=ek(ca), z=ek(ab)
xaybzc=eka(bc)ekb(ca)ekc(ab)=e0=1

Similarly, xb+c yc+a za+b=ek(b2c2)ek(c2a2)ek(a2b2)=e0=1

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