If $\frac{log x}{a^{2} + a b + b^{2}} = \frac{log y}{b^{2} + b c + c^{2}} = \frac{log z}{c^{2} + c a + a^{2}}$ $logx/(a^2+ab+b^2)=log y/(b^2+bc+c^2)=log z/(c^2+ca+a^2)$ then find $x^{a - b} \cdot y^{b - c} \cdot z^{c - a} =$ $x^(a-b)* y^(b-c)*z^(c-a)=$ ?

Question

If $\frac{log x}{a^{2} + a b + b^{2}} = \frac{log y}{b^{2} + b c + c^{2}} = \frac{log z}{c^{2} + c a + a^{2}}$ $logx/(a^2+ab+b^2)=log y/(b^2+bc+c^2)=log z/(c^2+ca+a^2)$ then find $x^{a - b} \cdot y^{b - c} \cdot z^{c - a} =$ $x^(a-b)* y^(b-c)*z^(c-a)=$ ?

2 Answers

Impact of this question

12896 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-08-11T13:43:02

Let
$\frac{log x}{a^{2} + a b + b^{2}} = \frac{log y}{b^{2} + b c + c^{2}} = \frac{log z}{c^{2} + c a + a^{2}} = k$ $logx/(a^2+ab+b^2)=log y/(b^2+bc+c^2)=log z/(c^2+ca+a^2)=k$

so
$log x = k (a^{2} + a b + b^{2})$ $logx=k(a^2+ab+b^2)$

$log y = k (b^{2} + b c + c^{2})$ $log y=k(b^2+bc+c^2)$

$log z = k (c^{2} + c a + a^{2})$ $log z=k(c^2+ca+a^2)$

Now

$log [x^{a - b} \cdot y^{b - c} \cdot z^{c - a}]$ $log[x^(a-b)* y^(b-c)*z^(c-a)]$

$= (a - b) log x + (b - c) log y + (c - a) log z$ $=(a-b)logx+ (b-c)logy+(c-a)logz$

$= (a - b) \times k (a^{2} + a b + b^{2}) + (b - c) l \times k (b^{2} + b c + c^{2}) + (c - a) l \times k (c^{2} + c a + a^{2})$ $=(a-b)xxk(a^2+ab+b^2)+ (b-c)lxxk(b^2+bc+c^2)+(c-a)lxxk(c^2+ca+a^2)$

$= k (a^{3} - b^{3} + b^{3} - c^{3} + c^{3} - a^{3}) = k \times 0 = 0 = log 1$ $=k(a^3-b^3+b^3-c^3+c^3-a^3)=kxx0=0=log1$

Hence

$x^{a - b} \cdot y^{b - c} \cdot z^{c - a} = 1$ $x^(a-b)* y^(b-c)*z^(c-a)=1$

Answer 2 · 2018-08-11T13:59:10

$x^{a - b} \cdot y^{b - c} \cdot z^{c - a} = 1$ $x^(a-b)*y^(b-c)*z^(c-a) =1$

Explanation:

We know that,

$(1) {log}_{z} A = P \Leftrightarrow A = z^{P}$ $(1)log_zA=P<=>A=z^P$

We take ,the base of log as $10$ $10$

Let ,

$\frac{log x}{a^{2} + a b + b^{2}} = \frac{log y}{b^{2} + b c + c^{2}} = \frac{log z}{c^{2} + c a + a^{2}} = k$ $logx/(a^2+ab+b^2)=logy/(b^2+bc+c^2)=logz/(c^2+ca+a^2)=k$

So ,

$log x = k (a^{2} + a b + b^{2}) \Rightarrow x = 10^{k (a^{2} + a b + b^{2})}$ $logx=k(a^2+ab+b^2)=>x=10^(k(a^2+ab+b^2))$

$log y = k (b^{2} + b c + c^{2}) \Rightarrow y = 10^{k (b^{2} + b c + c^{2})}$ $logy=k(b^2+bc+c^2)=>y=10^(k(b^2+bc+c^2)$

$log z = k (c^{2} + c a + a^{2}) \Rightarrow z = 10^{k (c^{2} + c a + a^{2})}$ $logz=k(c^2+ca+a^2)=>z=10^(k(c^2+ca+a^2)$

Now ,

$x^{a - b} = 10^{k (a - b) (a^{2} + a b + b^{2})} = 10^{k (a^{3} - b^{3})}$ $x^(a-b)=10^(k(a-b)(a^2+ab+b^2))=10^(k(a^3-b^3)$

$y^{b - c} = 10^{k (b - c) (b^{2} + b c + c^{2})} = 10^{k (b^{3} - c^{3})}$ $y^(b-c)=10^(k(b-c)(b^2+bc+c^2))=10^(k(b^3-c^3)$

$z^{c - a} = 10^{k (c - a) (c^{2} + c a + a^{2})} = 10^{k (c^{3} - a^{3})}$ $z^(c-a)=10^(k(c-a)(c^2+ca+a^2))=10^(k(c^3-a^3)$

Hence ,

$x^{a - b} \cdot y^{b - c} \cdot z^{c - a} = 10^{k (a^{3} - b^{3})} \cdot 10^{k (b^{3} - c^{3})} \cdot 10^{k (c^{3} - a^{3})}$ $x^(a-b)*y^(b-c)*z^(c-a) =10^(k(a^3-b^3) )*10^(k(b^3-c^3))*10^(k(c^3-a^3)$

$x^{a - b} \cdot y^{b - c} \cdot z^{c - a} = 10^{(k (a^{3} - b^{3})) + (k (b^{3} - c^{3})) + (k (c^{3} - a^{3})}$ $x^(a-b)*y^(b-c)*z^(c-a) =10^((k(a^3-b^3) )+(k(b^3-c^3))+(k(c^3-a^3)$

$x^{a - b} \cdot y^{b - c} \cdot z^{c - a} = 10^{k (a^{3} - b^{3} + b^{3} - c^{3} + c^{3} - a^{3})}$ $x^(a-b)*y^(b-c)*z^(c-a) =10^(k(a^3-b^3+b^3-c^3+c^3-a^3)$

$x^{a - b} \cdot y^{b - c} \cdot z^{c - a} = 10^{k (0)} = 10^{0} = 1$ $x^(a-b)*y^(b-c)*z^(c-a) =10^(k(0)) =10^0=1$

$x^{a - b} \cdot y^{b - c} \cdot z^{c - a} = 1$ $x^(a-b)*y^(b-c)*z^(c-a) =1$

Science

Math

Humanities

... and beyond

If $\frac{log x}{a^{2} + a b + b^{2}} = \frac{log y}{b^{2} + b c + c^{2}} = \frac{log z}{c^{2} + c a + a^{2}}$ $logx/(a^2+ab+b^2)=log y/(b^2+bc+c^2)=log z/(c^2+ca+a^2)$ then find $x^{a - b} \cdot y^{b - c} \cdot z^{c - a} =$ $x^(a-b)* y^(b-c)*z^(c-a)=$ ?

2 Answers

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

If logx/(a^2+ab+b^2)=log y/(b^2+bc+c^2)=log z/(c^2+ca+a^2)logxa2+ab+b2=logyb2+bc+c2=logzc2+ca+a2logx/(a^2+ab+b^2)=log y/(b^2+bc+c^2)=log z/(c^2+ca+a^2) then find x^(a-b)* y^(b-c)*z^(c-a)=xa−b⋅yb−c⋅zc−a=x^(a-b)* y^(b-c)*z^(c-a)=?

2 Answers

Explanation:

Related questions

Impact of this question

If $\frac{log x}{a^{2} + a b + b^{2}} = \frac{log y}{b^{2} + b c + c^{2}} = \frac{log z}{c^{2} + c a + a^{2}}$ $logx/(a^2+ab+b^2)=log y/(b^2+bc+c^2)=log z/(c^2+ca+a^2)$ then find $x^{a - b} \cdot y^{b - c} \cdot z^{c - a} =$ $x^(a-b)* y^(b-c)*z^(c-a)=$ ?