Question

If $a,b,c$ are distinct positive numbers, each different from 1, such that $[{\log }_{b}a{\log }_{c}a-{\log }_{a}a]+[{\log }_{a}b{\log }_{c}b-{\log }_{b}b]+[{\log }_{a}c{\log }_{b}c-{\log }_{c}c]=0,thenabc=$

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is A

$1$

Explanation for correct option:

Find the value of $abc$:

Given, $[{\log }_{b}a{\log }_{c}a-{\log }_{a}a]+[{\log }_{a}b{\log }_{c}b-{\log }_{b}b]+[{\log }_{a}c{\log }_{b}c-{\log }_{c}c]=0$

we know that, $\mathit{l}\mathit{o}{\mathit{g}}_{\mathbf{n}}\mathit{m}\mathbf{=}\frac{\mathbf{l}\mathbf{o}\mathbf{g}\mathbf{m}}{\mathbf{l}\mathbf{o}\mathbf{g}\mathbf{n}}$

using the above property in the given equation

$\begin{array}{rcl}(\frac{\log a}{\log b}\times \frac{\log a}{\log c}-\frac{\log a}{\log a})+(\frac{\log b}{\log a}\times \frac{\log b}{\log c}-\frac{\log b}{\log b})+(\frac{\log c}{\log a}\times \frac{\log c}{\log b}-\frac{\log c}{\log c})& =& 0\end{array}$

$\Rightarrow$ $\begin{array}{rcl}& & \begin{array}{ccl}[\frac{{\left(\log a\right)}^2}{\log b\times \log c}-1]+[\frac{{\left(\log b\right)}^2}{\log a\times \log c}-1]+[\frac{{\left(\log c\right)}^2}{\log a\times \log b}-1]& =& 0\end{array}\end{array}$

$\Rightarrow$ $\begin{array}{rcl}\frac{{\left(\log a\right)}^3+{\left(\log b\right)}^3+{\left(\log c\right)}^3}{\log a\times \log b\times \log c}& =& 3\end{array}$

$\Rightarrow$ $\begin{array}{rcl}{\left(\log a\right)}^3+{\left(\log b\right)}^3+{\left(\log c\right)}^3& =& 3\times \log a\times \log b\times \log c\end{array}$

We know that, if $\begin{array}{rcl}{p}^3+q^3+r^3& =& 3pqr\end{array}$

Then $\begin{array}{rcl}p+q+r& =& 0\end{array}$ using this property,

$\begin{array}{rcl}\left({\log }_{10}a\right)+\left({\log }_{10}b\right)+\left({\log }_{10}c\right)& =& 0\end{array}$

$\Rightarrow$ $\begin{array}{rcl}{\log }_{10}abc& =& 0\end{array}$

$\begin{array}{rcl}\mathbf{\therefore }\mathit{a}\mathit{b}\mathit{c}& \mathbf{=}& \mathbf{1}\end{array}$

Hence, correct option is $\left(A\right)$