Question

$\text{If }\dfrac{\log x}{l+m-2n}=\dfrac{\log y}{m+n-2l}=\dfrac{\log z}{n+l-2m},\text{ then find }{{x}^{2}}{{y}^{2}}{{z}^{2}}.$

Answer 1

Hint: First of all equate the given expression to a constant k. Now, find the value of log x, log y, and log z in terms of this constant and add them. Now, use log m + log n + log q = log mnq and log a = b then ${{10}^{b}}$ to find the value of xyz. Now, square both the sides to get the value of ${{x}^{2}}{{y}^{2}}{{z}^{2}}$.

Complete step-by-step answer:

In this question, we are given that $\dfrac{\log x}{l+m-2n}=\dfrac{\log y}{m+n-2l}=\dfrac{\log z}{n+l-2m}$, we have to find the value of ${{x}^{2}}{{y}^{2}}{{z}^{2}}$. Let us consider the expression given in the question and equate it to a constant k as follows:
$\dfrac{\log x}{l+m-2n}=\dfrac{\log y}{m+n-2l}=\dfrac{\log z}{n+l-2m}=k$
By separating the above equation into parts, we get,
$\dfrac{\log x}{l+m-2n}=k$
$\log x=k\left( l+m-2n \right)....\left( i \right)$
Similarly, we also get,
$\dfrac{\log y}{m+n-2l}=k$
$\log y=k\left( m+n-2l \right)....\left( ii \right)$
$\dfrac{\log z}{n+l-2m}=k$
$\log z=k\left( n+l-2m \right)....\left( iii \right)$
Now, by adding equation (i), (ii), and (iii), we get,
$\log x+\log y+\log z=k\left( l+m-2n \right)+k\left( m+n-2l \right)+k\left( n+l-2m \right)$
By taking out k common from RHS of the above equation, we get,
$\log x+\log y+\log z=k\left[ l+m-2n+m+n-2l+n+l-2m \right]$
By rearranging the terms of the RHS of the above equation, we get,
$\log x+\log y+\log z=k\left[ \left( l+l-2l \right)+\left( m+m-2m \right)+\left( n+n-2n \right) \right]$
$\log x+\log y+\log z=k\left[ 0+0+0 \right]$
$\log x+\log y+\log z=0$
We know that, log m + log n + log q = log (mnq). By using this in the above equation, we get,
log xyz = 0
We know that when log a = b, then $a={{10}^{b}}$. By using this, we get,
$xyz={{10}^{0}}$
We know that any number raised to 0 is equal to 1. By using this, we get, xyz = 1.
By squaring both the sides of the above equation, we get,
${{\left( xyz \right)}^{2}}={{1}^{2}}$
We know that, ${{\left( abc \right)}^{x}}={{a}^{x}}.{{b}^{x}}.{{c}^{x}}$. By using this, we get,
${{x}^{2}}.{{y}^{2}}.{{z}^{2}}=1$
Hence, we have got the value of ${{x}^{2}}{{y}^{2}}{{z}^{2}}$ as 1.

Note: In these types of questions, students often try to find the value of x, y, and z individually and then multiply them to get xyz which is also correct but as we know the property of the sum of the logarithm, that is $\log {{m}_{1}}+\log {{m}_{2}}+\log {{m}_{3}}....\log {{m}_{n}}=\log {{m}_{1}}{{m}_{2}}{{m}_{3}}....{{m}_{n}}$. So, whenever the product of the variables and log are simultaneously coming in the question, it is always advisable to take the sum of the log from which the product can be obtained as we have done in the above solution. And in this way, we can get the required expression very easily.