Q.289. Logarithm

Question 289.
If [logx/(a - b)] = [logy/(b - c)] = [logz/(c - a)], then prove that xyz = 1 and x^c y^a z^b = 1.

Answer 289.
Let logx / (a - b) = logy / (b - c) = logz / (c - a) = k

=> logx = k (a - b), log y = k (b - c) and logz = k (c - a)
=> logx + logy + logz = k (a - b + b - c + c - a) = 0
=> log (xyz) = 0
=> xyz = 1.

x^c y^a z^b
= Antilog log (x^c y^a z^b)
= Antilog [clogx + alogy + blogz)
= Antilog [ck(a - b) + ak(b - c) + bk(c - a)]
= Antilog 0
= 1.

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