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+1 vote
17.7k views
in Logarithm by (24.0k points)
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If x = logabc, y = logbca, z = logcab, then

(a) xyz = x + y + z + 2 

(b) xyz = x + y + z + 1 

(c) x + y + z = 1 

(d) xyz = 1

1 Answer

+2 votes
by (23.5k points)
selected by
 
Best answer

(a) x + y + z + 2 = xyz.

x = logabc ⇒ ax = bc ⇒ ax + 1 = abc ⇒ a = (abc)1/(x +1) 

Similarly, b = (abc)1/(y + 1), c = (abc)1/(z + 1)

∴ abc = \((abc)^{\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}}\)

⇒ \({\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}}\) = 1

⇒ (y + 1) (z + 1) + (x + 1) (z + 1) + (x + 1) (y + 1) = (x + 1) (y + 1) (z + 1)

⇒ yz + y + z + 1 + xz + x + z + 1 + xy + y + x + 1 = xyz + xy + yz + zx + x + y + z + 1

⇒ x + y + z + 2 = xyz.

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