Question

Solve the following, if given that : $\dfrac{{\log x}}{{\left( {a - b} \right)}} = \dfrac{{\log y}}{{\left( {b - c} \right)}} = \dfrac{{\log z}}{{\left( {c - a} \right)}}$. Then, find the value of $xyz$.
(a) $- 1$
(b) Cannot be determined
(c) $1$
(d) $0$

Answer 1

Hint : The given problem revolves around the concepts of logarithms. To find the value we will first assume the given condition as any variable, say, ‘$l$’. After assuming, taking the solution for each term then taking the log for required solution, using the certain rules of indices, logarithms formulae for multiplication, addition, etc. to get the desired value.

Complete step-by-step answer :
Since, we have given the equation that
$\dfrac{{\log x}}{{\left( {a - b} \right)}}{\text{ }} = {\text{ }}\dfrac{{\log y}}{{\left( {b - c} \right)}}{\text{ }} = {\text{ }}\dfrac{{\log z}}{{\left( {c - a} \right)}}$
So, let us assume that ‘$l$’ equates the given equations, we get
$\dfrac{{\log x}}{{\left( {a - b} \right)}}{\text{ }} = {\text{ }}\dfrac{{\log y}}{{\left( {b - c} \right)}}{\text{ }} = {\text{ }}\dfrac{{\log z}}{{\left( {c - a} \right)}}{\text{ }} = {\text{ }}l$
Hence, the each term becomes,
$\dfrac{{\log x}}{{\left( {a - b} \right)}}{\text{ }} = {\text{ }}l$
Solving the equation mathematically, we get
$\log x{\text{ }} = {\text{ }}\left( {a - b} \right)l$ … ($1$)
Similarly,
$\dfrac{{\log y}}{{\left( {b - c} \right)}}{\text{ }} = {\text{ }}l$
$\log y{\text{ }} = {\text{ }}\left( {b - c} \right)l$ … ($2$)
And,
$\dfrac{{\log z}}{{\left( {c - a} \right)}}{\text{ }} = {\text{ }}l$
$\log z{\text{ }} = {\text{ }}\left( {c - a} \right)l$ … ($3$)
Now, considering the given expression for which we need to find the value, that is
$= {\text{ }}xyz$
Taking log, the expression becomes
$= {\text{ }}\log \left( {xyz} \right)$
By using the logarithmic rule for multiplication that is$\log \left( {a.b} \right){\text{ }} = {\text{ }}\log a{\text{ }} + {\text{ }}\log b$, we get
$\log xyz{\text{ }} = {\text{ }}\log x{\text{ }} + {\text{ }}\log y{\text{ }} + {\text{ }}\log z$
From ($1$), ($2$), and ($3$)
Substituting in the above equation, we get
$\log \left( {xyz} \right){\text{ }} = {\text{ }}\left( {a - b} \right)l{\text{ }} + {\text{ }}\left( {b - c} \right)l{\text{ }} + {\text{ }}\left( {c - a} \right)l$
Solving the equation mathematically, we get
$\log \left( {xyz} \right){\text{ }} = {\text{ }}l\left( 0 \right)$
$\log \left( {xyz} \right){\text{ }} = {\text{ }}0$
We know that,
According to rule for indices of logarithms that is,
When,
$\log x{\text{ }} = {\text{ }}y$
$x{\text{ }} = {\text{ }}{e^y}$
As a result, the equation becomes
$xyz{\text{ }} = {\text{ }}{e^0}$
Hence, we know that${\left( {anything} \right)^0}{\text{ }} = {\text{ }}1$, we get
$xyz{\text{ }} = {\text{ }}1$
$\therefore \Rightarrow$The option (c) is correct!
So, the correct answer is “Option c”.

Note : In this case, assumption of any variable for the given condition is necessary. One must know the rules of the indices, rules of logarithms that is $\log \left( {a.b} \right){\text{ }} = {\text{ }}\log a{\text{ }} + {\text{ }}\log b$, ${\left( {anything} \right)^0}{\text{ }} = {\text{ }}1$, etc., so as to be sure of our final answer.