Question

If ${\left( {abc} \right)^{a + b + c}} = 1$ where $abc \ne 1$ and ${\log _3}\left( {abc} \right) = \alpha $ , then ${\log _{abc}}\left( {{a^3} + {b^3} + {c^3}} \right) = $
A) $\dfrac{{1 - \alpha }}{\alpha }$
B) $\dfrac{\alpha }{{1 + \alpha }}$
C) $\dfrac{{1 + \alpha }}{\alpha }$
D) $\dfrac{1}{\alpha }$

Answer 1

Hint:
From the given equation using the properties of logarithm, we can find the value of $a + b + c$. Then we can simplify the 2nd equation the property of logarithms that ${\log _a}b = \dfrac{{\log b}}{{\log a}}$ . Then we can take the expansion ${\left( {a + b + c} \right)^3} = {a^3} + {b^3} + {c^3} + 3\left( {a + b} \right)\left( {b + c} \right)\left( {c + a} \right)$ and rearrange and simply using the relations we derived. Then we can take the required logarithm to the base abc. Then by simplification and substitution, we can obtain the required solution.

Complete step by step solution:
We are given that ${\left( {abc} \right)^{a + b + c}} = 1$ and $abc \ne 1$ .
We can take the logarithm on both sides.
 $ \Rightarrow \log {\left( {abc} \right)^{a + b + c}} = \log 1$
We know that $\log 1 = 0$ and $\log {a^b} = b\log a$
 $ \Rightarrow \left( {a + b + c} \right)\log \left( {abc} \right) = 0$
It is given that, $abc \ne 1$ ,
 $ \Rightarrow \log \left( {abc} \right) \ne 0$
 $ \Rightarrow \left( {a + b + c} \right) = 0$ …. (1)
So, we can write the sum of 2 terms as the negative of the third term.
 $ \Rightarrow a + b = - c$ … (2)
 $ \Rightarrow a + c = - b$ … (3)
 $ \Rightarrow b + c = - a$ … (4)
We are given that ${\log _3}\left( {abc} \right) = \alpha $
We know that ${\log _a}b = \dfrac{{\log b}}{{\log a}}$ . So, we can write the given equation as,
 $ \Rightarrow \dfrac{{\log \left( {abc} \right)}}{{\log 3}} = \alpha $ … (5)
We need to find the value of ${\log _{abc}}\left( {{a^3} + {b^3} + {c^3}} \right)$
We know that the terms inside the log are the terms in the cubic expansion of 3 numbers. It is given by,
 ${\left( {a + b + c} \right)^3} = {a^3} + {b^3} + {c^3} + 3\left( {a + b} \right)\left( {b + c} \right)\left( {c + a} \right)$
On rearranging, we get,
 $ \Rightarrow {a^3} + {b^3} + {c^3} = {\left( {a + b + c} \right)^3} - 3\left( {a + b} \right)\left( {b + c} \right)\left( {c + a} \right)$
 On substituting equations (1), (2), (3) and (4), we get,
 $ \Rightarrow {a^3} + {b^3} + {c^3} = {\left( 0 \right)^3} - 3\left( { - c} \right)\left( { - a} \right)\left( { - b} \right)$
On simplification of the negative signs we get,
 $ \Rightarrow {a^3} + {b^3} + {c^3} = 3abc$
Now we can take ${\log _{abc}}$ on both sides.
 $ \Rightarrow {\log _{abc}}\left( {{a^3} + {b^3} + {c^3}} \right) = {\log _{abc}}\left( {3abc} \right)$
We know that $\log ab = \log a + \log b$ . Applying this on RHS, we get,
 $ \Rightarrow {\log _{abc}}\left( {{a^3} + {b^3} + {c^3}} \right) = {\log _{abc}}3 + {\log _{abc}}\left( {abc} \right)$
We know that ${\log _a}b = \dfrac{{\log b}}{{\log a}}$ and ${\log _a}a = 1$ .
 \[ \Rightarrow {\log _{abc}}\left( {{a^3} + {b^3} + {c^3}} \right) = \dfrac{{\log 3}}{{\log \left( {abc} \right)}} + 1\]
Now we can substitute equation (5).
 \[ \Rightarrow {\log _{abc}}\left( {{a^3} + {b^3} + {c^3}} \right) = \dfrac{1}{\alpha } + 1\]
On taking the LCM of the denominator of the RHS, we get,
 \[ \Rightarrow {\log _{abc}}\left( {{a^3} + {b^3} + {c^3}} \right) = \dfrac{{1 + \alpha }}{\alpha }\]

Therefore, the required value is \[\dfrac{{1 + \alpha }}{\alpha }\] .
So, the correct answer is option C.

Note:
The properties of logarithm used in this problem are,
$\log {a^b} = b\log a$
${\log _a}b = \dfrac{{\log b}}{{\log a}}$ .
We know that logarithm is defined for only positive values and is not defined for negative values and at zero. So before taking the logarithm of the expression, we must make sure that both the sides are positive. We can only expand the multiplication and division inside the logarithm and we cannot expand the addition or subtraction inside the log.