Question

if logx:logy:logz=(y-z):(z-x):(x-y) then 1=

Answer 1

If logx:logy:logz=(y-z):(z-x):(x-y) then xyz = 1

Step-by-step explanation:

Given

$\log x:\log y:\log z=(y-z):(z-x):(x-y)$

Let the base of log be a

If the common ratio is k

Then

$\log_a x=(y-z)k$

$\implies x=a^{(y-z)k}$             (∵ By definition of log if $\log_a x=y\implies x=a^y$)

$\log_a y=(z-x)k$

$\implies y=a^{(z-x)k}$

$\log_a z=(x-y)k$

$\implies z=a^{(x-y)k}$

Thus,

$xyz=a^{(y-z)k}\times a^{(z-x)k\times a^{(x-y)k}$

$\implies xyz=a^{k(y-z+z-x+x-y)}$        (∵ $a^m\times a^n=a^{m+n}$)

$\implies xyz=a^0$   (∵ $a^0=1$)

$\implies xyz=1$

Hope this answer is helpful.

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