Given log⁡xb−c=log⁡yc−a=log⁡za−b show that xb+c−a⋅yc+a−b⋅za+b−c=1

Question

Given log⁡xb−c=log⁡yc−a=log⁡za−b show that  xb+c−a⋅yc+a−b⋅za+b−c=1

maul124uk · Accepted Answer

We have  log⁡xb−c=log⁡yc−a=log⁡za−b=t  This gives us  x=et(b−c),y=et(c−a)  and  z=et(a−b)  Hence,  xb+c−a⋅yc+a−b⋅za+b+c=e((b−c)(b+c−a)+(c−a)(c+a−b)+(a−b)(a+b−c))  =et(b2−c2−ab+ac+c2−a2−bc+ba+a2−b2−ac+bc)=e0=1

Karen Robbins · Accepted Answer

If you want to use your equations, here is a method.  Multiplying the equations together, we obtain:  xc−aya−bzb−c=yb−czc−axa−b  which gives after reordering:  Therefore it suffices to show that xaybzc=1  Your first and third equations give y=xc−ab−c,z=xa−bb−c This gives us:  xaybzc=xaxc−ab−c×bxa−bb−c×c=xa+bc−ba+ca−bcb−c=xa−a=x0=1  QED

Vasquez · Accepted Answer

Given: log⁡xb−c=log⁡yc−a=log⁡za−b=λ we have: x=eλ(b+c),y=eλ(c−a),z=eλ(a−b) hence: xb+c−a⋅yc+a−b⋅za+b−c=exp⁡(λ⋅∑cyc(b2−c2−a(b−c))) =exp⁡(0)=1

Given \frac{\log x}{b-c}=\frac{\log y}{c-a}=\frac{\log z}{a-b} show that x^{b+c-a}\cdot y^{c+a-b}\cdot

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