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KCET · Maths · Basic of Mathematics

If \(\frac{\log x}{b-c}=\frac{\log y}{c-a}=\frac{\log z}{a-b}\), then the value of \(x^{b+c} \cdot y^{c+a} \cdot z^{a+b}\) is

  1. A 1
  2. B 2
  3. C 0
  4. D \(-1\)
Verified Solution

Answer & Solution

Correct Answer

(A) 1

Step-by-step Solution

Detailed explanation

Given, \(\frac{\log x}{b-c}=\frac{\log y}{c-a}=\frac{\log z}{a-b}=k\) (say)
\(\ldots(i)\) \(\Rightarrow \quad x=e^{k(b-c)}, y=e^{k(c-a)}, z=e^{k(a-b)}\)
Now, \(\quad x^{b+c} \cdot y^{c+a} \cdot z^{a+b}\) \(=e^{k(b-c)(b+c)} \cdot e^{k(c+a)(c-a)} \cdot e^{k(a+b)(a-b)}\)
\[
\begin{aligned}
&=e^{k\left(b^{2}-c^{2}\right)} \cdot e^{k\left(c^{2}-a^{2}\right)} \cdot e^{k\left(a^{2}-b^{2}\right)} \\
&=e^{k\left(b^{2}-c^{2}+c^{2}-a^{2}+a^{2}-b^{2}\right)}=e^{k \cdot 0}=1
\end{aligned}
\]