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Question

If logxbc=logyca=logzab, then which of the following is/are true?

A
xyz=1
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B
xaybzc=1
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C
xb+cyc+aza+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+cyc+aza+b=1
D xyz=xaybzc
logkxbc=logkyca=logkzab=p (let)logkxbc=px=k(p(bc))Similarly, y=k p(ca) and z=k p(ab)xyz=k p(bc+ca+ab)=1xaybzc=k p(abac+bcab+acbc)=1xb+cyc+aza+b=k p(b2c2+c2a2+a2b2)=1

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