Question

If $a,b,c$ are distinct positive numbers each being different form 1 such that $({\log }_{b}a.{\log }_{c}a-{\log }_{a}a)+({\log }_{a}b.{\log }_{c}b-{\log }_{b}b)+({\log }_{a}c.{\log }_{b}c-{\log }_{c}c)=0$, then $abc$is equal to

• A. $0$
• B. $e$
• C. $1$
• D. $2$
• E. $5$

Answer 1

The correct option is C

$1$

Finding the value of :
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, $\left[{\log }_n\left(m\right)=\frac{\mathrm{logm}}{\mathrm{logn}}\right]$

Using the above property then given equation

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

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

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

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

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 $\mathbf{\left(}\mathit{C}\mathbf{\right)}\mathbf{.}$