If frac{log x}{b-c}=frac{log y}{c-a}=frac{log | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given,$\frac{\log x}{b-c}=\frac{\log y}{c-a}=\frac{\log z}{a-b}=k$ (say). From this,we have $\log x = k(b-c)$,$\log y = k(c-a)$,and $\log z = k(a-

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(A) Given,$\frac{\log x}{b-c}=\frac{\log y}{c-a}=\frac{\log z}{a-b}=k$ (say). From this,we have $\log x = k(b-c)$,$\log y = k(c-a)$,and $\log z = k(a-b)$. This implies $x = e^{k(b-c)}$,$y = e^{k(c-a)}$,and $z = e^{k(a-b)}$. Now,consider the expression $E = x^{b+c} \cdot y^{c+a} \cdot z^{a+b}$. Substituting the values,$E = (e^{k(b-c)})^{b+c} \cdot (e^{k(c-a)})^{c+a} \cdot (e^{k(a-b)})^{a+b}$. Using the property $(e^m)^n = e^{mn}$,we get $E = e^{k(b^2-c^2)} \cdot e^{k(c^2-a^2)} \cdot e^{k(a^2-b^2)}$. Combining the exponents,$E = e^{k(b^2-c^2+c^2-a^2+a^2-b^2)}$. Since the sum in the exponent is $0$,$E = e^{k \cdot 0} = e^0 = 1$.

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

Similar Questions

The solution to the equation ${\log _7}{\log _5}(\sqrt {{x^2} + 5 + x} ) = 0$ is:

If $\frac{\log x}{b - c} = \frac{\log y}{c - a} = \frac{\log z}{a - b}$,then which of the following is true?

If $\log _{0.3}(x-1) < \log _{0.09}(x-1),$ then $x$ lies in the interval

If $a, b, c \neq 0$ and belong to the set $\{1, 2, 3, \ldots, 9\}$,then $\log _{10}\left(\frac{a+10 b+10^2 c}{10^{-4} a+10^{-3} b+10^{-2} c}\right)$ is equal to

If $\cosh ^{-1}\left(\frac{5}{3}\right)+\sinh ^{-1}\left(\frac{3}{4}\right)=k$,then $e^k=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module