If x = logabc, y = logbca, z = logc ab, then -Maths 9th

If x = logabc, y = logbca, z = logc ab, then -Maths 9th
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(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 = \((abc)^{rac{1}{x+1}+rac{1}{y+1}+rac{1}{z+1}}\)⇒ \({rac{1}{x+1}+rac{1}{y+1}+rac{1}{z+1}}\) = 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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Last Answer : Join AC and QP, also it is given that AQ || CP ∴ △ACQ and △APQ are on the same base AQ and lie between the same parallels AQ || CP. ∴ ar(△ACQ) = ar(△APQ) or ar(△ABC) + ar(△ABQ) = ar(△BPQ) + ar(△ABQ) or ar(△ABC) = ar( △BPQ) or 1/2 ar(||gm ABCD) = 1/2 ar(||gm PBQR) or ar(||gm ABCD) = ar(||gm PBQR)