Question

if log^x÷a(b-c)=log^y÷b(c-a)=log^z÷c(a-b) then show that xyz=1​

Answer 1

Step-by-step explanation:

given that

$\frac{ log(x) }{ a(b - c)} = \frac{ log(y) }{b(c - a)} = \frac{ log(z) }{c(a - b)} = k \\ so \: log(x ) = ka(b - c) \\ log(y) = kb(c - a ) \\ log(z) = kc(a - b) \\ log(x) + log(y) + log(z) = ka(b - c) + kb(c - a) + kc(a - b) \\ log(xyz) = k{a(b - c) + b(c - a) + c(a - b)} \\ log(xyz) = k \times 0 \\ log(xyz) = 0 \\ so \\ xyz = 1 \\ \\ from \: log \: property \\ i \: hope \: t \: will \: help \: you$

✌