Question

If ${\log }_{b}a.{\log }_{c}a+{\log }_{a}b.{\log }_{c}b+{\log }_{a}c.{\log }_{b}c=3$ (where a, b, c are different positive real numbers ≠ 1) then find the value of $abc$ .

Answer 1

Hint: A logarithm, of a base b, is the power to which the base needs to be raised to yield a given number. We know that ${\log }_{b}a={\displaystyle \frac{\log a}{\log b}}$ , so first convert the given logarithmic terms with bases to this form using this conversion. And then solve the remaining solution referring to the below mentioned formula.
Formulas used:
 ${\log }_{b}a={\displaystyle \frac{\log a}{\log b}}$
If $x+y+z=0$ , then $x^3+y^3+z^3=3xyz$ and vice-versa.

Complete step-by-step answer:
We are given a logarithmic equation ${\log }_{b}a.{\log }_{c}a+{\log }_{a}b.{\log }_{c}b+{\log }_{a}c.{\log }_{b}c=3$ where a, b, c are different positive real numbers ≠ 1.
We have to find the value of $abc$ .
As we already know that ${\log }_{b}a={\displaystyle \frac{\log a}{\log b}}$ .
Therefore, Using the above conversion we are converting the logarithmic terms present in the given equation into this fractional form.
 ${\log }_{c}a={\displaystyle \frac{\log a}{\log c}}$
 ${\log }_{a}b={\displaystyle \frac{\log b}{\log a}}$
 ${\log }_{c}b={\displaystyle \frac{\log b}{\log c}}$
 ${\log }_{a}c={\displaystyle \frac{\log c}{\log a}}$
 ${\log }_{b}c={\displaystyle \frac{\log c}{\log b}}$
On substituting all the obtained fractional terms in ${\log }_{b}a.{\log }_{c}a+{\log }_{a}b.{\log }_{c}b+{\log }_{a}c.{\log }_{b}c=3$ , we get
 $\left({\displaystyle \frac{\log a}{\log b}}\times {\displaystyle \frac{\log a}{\log c}}\right)+\left({\displaystyle \frac{\log b}{\log a}}\times {\displaystyle \frac{\log b}{\log c}}\right)+\left({\displaystyle \frac{\log c}{\log a}}\times {\displaystyle \frac{\log c}{\log b}}\right)=3$
 $\Rightarrow {\displaystyle \frac{{\left(\log a\right)}^2}{\log b.\log c}}+{\displaystyle \frac{{\left(\log b\right)}^2}{\log a.\log c}}+{\displaystyle \frac{{\left(\log c\right)}^2}{\log a.\log b}}=3$
Take out the LCM and convert the above left hand side into a single fraction, the LCM is $\log a.\log b.\log c$
 $\Rightarrow {\displaystyle \frac{\log a\times {\left(\log a\right)}^2+\log b\times {\left(\log b\right)}^2+\log c\times {\left(\log c\right)}^2}{\log a.\log b.\log c}}=3$
 $\Rightarrow {\displaystyle \frac{{\left(\log a\right)}^3+{\left(\log b\right)}^3+{\left(\log c\right)}^3}{\log a.\log b.\log c}}=3$
On cross multiplication, we get
 $\Rightarrow {\left(\log a\right)}^3+{\left(\log b\right)}^3+{\left(\log c\right)}^3=3\times \left(\log a.\log b.\log c\right)=3\log a.\log b.\log c$
As we can see, the above equation is in the form of $x^3+y^3+z^3=3xyz$ , where x is $\log a$ , y is $\log b$ and z is $\log c$
Therefore, $x+y+z$ must be equal to zero which means $\log a+\log b+\log c=0$
We know that $\log a+\log b$ is equal to $\log ab$
Therefore, $\log a+\log b+\log c=\log abc$
 $\log abc=0$
Sending the logarithm to the right hand side (as $\log abc$ is a common logarithm it will have a base 10)
 $abc={10}^0=1$ (Anything to the power zero is equal to 1)
Therefore, the value of $abc$ is 1.
So, the correct answer is “1”.

Note: We know that ${\log }_{b}a={\displaystyle \frac{\log a}{\log b}}$ , which can also be written as ${\displaystyle \frac{1}{\left({\displaystyle \frac{\log b}{\log a}}\right)}}={\displaystyle \frac{1}{{\log }_{a}b}}$ . And while finding the value of ${\log }_{b}a$ , confirm that b is always greater than zero and never equal 1; a must be a positive real number. If ${\log }_{b}a=k$ , then $a=b^k$