Question

If $\dfrac{\log x}{l+m-2n}=\dfrac{\log y}{m+n-2l}=\dfrac{\log z}{n+l-2m}$ , then the value of ${{x}^{2}}{{y}^{2}}{{z}^{2}}$ will be:
A. 0
B. 1
C. 2
D. 3

Answer 1

Hint: We can solve this question by finding separate values of x , y and z in terms of log by equating a given equation to any constant. Then by solving all the equations of the constant we will arrive at the value of xyz.

Complete step-by-step answer:
Given equation is
$\dfrac{\log x}{l+m-2n}=\dfrac{\log y}{m+n-2l}=\dfrac{\log z}{n+l-2m}$
We can let it equal to k
So we can write equation as
$\dfrac{\log x}{l+m-2n}=\dfrac{\log y}{m+n-2l}=\dfrac{\log z}{n+l-2m}=k$
Now we can write:
$\Rightarrow \dfrac{\log x}{l+m-2n}=k$
$\Rightarrow \log x=k(l+m-2n)$ ……………………………………(i)
$\Rightarrow \dfrac{\log y}{m+n-2l}=k$
$\Rightarrow \log y=k(m+n-2l)$ …………………………………………..(ii)
$\Rightarrow \dfrac{\log z}{n+l-2m}=k$
$\Rightarrow \log z=k(n+l-2m)$ ……………………………………………(iii)
On adding equation (i),(ii) and (iii)
$\Rightarrow \log x+\log y+\log z=k(l+m-2n)+k(m+n-2l)+k(n+l-2m)$
$\Rightarrow \log x+\log y+\log z=kl+km-2kn+km+kn-2kl+kn+kl-2km$
$\Rightarrow \log x+\log y+\log z=0$
$\Rightarrow \log (xyz)=0$ \left\\{ \because \log a+\log b+\log c=\log (abc) \right\\}
According to base change formula if we have ${{\log }_{a}}x=b$ then we can write $x={{a}^{b}}$
Hence we can write above equation as:
$\Rightarrow xyz={{10}^{0}}$
$\Rightarrow xyz=1$
On squaring both side
$\Rightarrow {{(xyz)}^{2}}={{1}^{2}}$
$\Rightarrow {{x}^{2}}{{y}^{2}}{{z}^{2}}=1$
Hence option B is correct.

Note:In this question we need to remember that all equations are equal. So we will let all equations equal to the same variable. Also if any terms has exponent 0 then the value of term is equal to 1 always.