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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 $ .

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Last updated date: 20th Jun 2024
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Answer
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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 = \dfrac{{\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 = \dfrac{{\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 = \dfrac{{\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 = \dfrac{{\log a}}{{\log c}} $
 $ {\log _a}b = \dfrac{{\log b}}{{\log a}} $
 $ {\log _c}b = \dfrac{{\log b}}{{\log c}} $
 $ {\log _a}c = \dfrac{{\log c}}{{\log a}} $
 $ {\log _b}c = \dfrac{{\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( {\dfrac{{\log a}}{{\log b}} \times \dfrac{{\log a}}{{\log c}}} \right) + \left( {\dfrac{{\log b}}{{\log a}} \times \dfrac{{\log b}}{{\log c}}} \right) + \left( {\dfrac{{\log c}}{{\log a}} \times \dfrac{{\log c}}{{\log b}}} \right) = 3 $
 $ \Rightarrow \dfrac{{{{\left( {\log a} \right)}^2}}}{{\log b.\log c}} + \dfrac{{{{\left( {\log b} \right)}^2}}}{{\log a.\log c}} + \dfrac{{{{\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 \dfrac{{\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 \dfrac{{{{\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 = \dfrac{{\log a}}{{\log b}} $ , which can also be written as $ \dfrac{1}{{\left( {\dfrac{{\log b}}{{\log a}}} \right)}} = \dfrac{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} $