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Question: If \(a^{x - 1} = bc,b^{y - 1} = ca,c^{z - 1} = ab,\) then \(\sum_{}^{}{(1/x) =}\)...

If ax−1=bc,by−1=ca,cz−1=ab,a^{x - 1} = bc,b^{y - 1} = ca,c^{z - 1} = ab, then ∑(1/x)=\sum_{}^{}{(1/x) =}

A

1

B

0

C

abc

D

None of these

Answer

1

Explanation

Solution

ax−1=bc⇒ax=abca^{x - 1} = bc \Rightarrow a^{x} = abc

∴\therefore ax=by=cz=abc=k(say)a^{x} = b^{y} = c^{z} = abc = k(\text{say})

⇒a=k1/x⇒1x=log⁡ka\Rightarrow a = k^{1/x} \Rightarrow \frac{1}{x} = \log_{k}a;

∴∑1x=log⁡ka+log⁡kb+log⁡kc=log⁡kabc=log⁡abcabc=1\sum_{}^{}{\frac{1}{x} = \log_{k}a + \log_{k}b + \log_{k}c} = \log_{k}abc = \log_{abc}abc = 1