If x=loga(bc),y=logb(ca) and z=logc(ab) , then which of the following equation is equal to 1?
A
x+y+z
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B
(1+x)−1+(1+y)−1+(1+z)−1
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C
xyz
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D
none of these
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Solution
The correct option is D(1+x)−1+(1+y)−1+(1+z)−1 x=logabc ⇒1+x=logb+logcloga+1⇒1+x=logb+logc+logaloga ...(1) y=logbac ⇒1+y=loga+logclogb+1⇒1+y==loga+logc+logblogb ...(2) z=logcab⇒1+z=loga+logblogc+1 ⇒1+z=loga+logb+logclogc ...(3) From (1), (2) & (3), we get (1+x)−1+(1+y)−1+(1+z)−1=loga+logb+logcloga+logb+logc=1 Ans: B