Question

If ${{\log }_{3}}\left( {{\log }_{3}}a \right)+{{\log }_{\dfrac{1}{3}}}\left( {{\log }_{\dfrac{1}{3}}}b \right)=1,$ then value of $a{{b}^{3}}$ is
(a) 9
(b) 3
(c) 1
(d) $\dfrac{1}{3}$

Answer 1

Hint: We start solving the problem by writing 1 as ${{\log }_{3}}3$. We then make use of results ${{\log }_{\dfrac{1}{p}}}q=-{{\log }_{p}}q$, $-{{\log }_{p}}q={{\log }_{p}}\left( \dfrac{1}{q} \right)$, ${{\log }_{d}}e-{{\log }_{d}}f={{\log }_{d}}\left( \dfrac{e}{f} \right)$ to proceed through the problem. We then make use of the fact that if ${{\log }_{a}}d={{\log }_{a}}e$, then we get $d=e$ to proceed further. We then make use of the results $\dfrac{{{\log }_{d}}e}{{{\log }_{d}}f}={{\log }_{f}}e$ and if ${{\log }_{d}}e=f$, then $e={{d}^{f}}$ to get the value of $a{{b}^{3}}$.

Complete step by step answer:
According to the problem, we are given that ${{\log }_{3}}\left( {{\log }_{3}}a \right)+{{\log }_{\dfrac{1}{3}}}\left( {{\log }_{\dfrac{1}{3}}}b \right)=1,$ we need to find the value of $a{{b}^{3}}$.
So, we have ${{\log }_{3}}\left( {{\log }_{3}}a \right)+{{\log }_{\dfrac{1}{3}}}\left( {{\log }_{\dfrac{1}{3}}}b \right)=1$ ---(1).
We know that ${{\log }_{p}}p=1$. Let us use this result in equation (1) by taking $p=3$.
$\Rightarrow {{\log }_{3}}\left( {{\log }_{3}}a \right)+{{\log }_{\dfrac{1}{3}}}\left( {{\log }_{\dfrac{1}{3}}}b \right)={{\log }_{3}}3$ ---(2).
We know that ${{\log }_{\dfrac{1}{p}}}q=-{{\log }_{p}}q$. Let us use this result in equation (2).
$\Rightarrow {{\log }_{3}}\left( {{\log }_{3}}a \right)-{{\log }_{3}}\left( -{{\log }_{3}}b \right)={{\log }_{3}}3$ ---(3).
We know that $-{{\log }_{p}}q={{\log }_{p}}\left( \dfrac{1}{q} \right)$. Let us use this result in equation (3).
$\Rightarrow {{\log }_{3}}\left( {{\log }_{3}}a \right)-{{\log }_{3}}\left( {{\log }_{3}}\dfrac{1}{b} \right)={{\log }_{3}}3$ ---(4).
We know that ${{\log }_{d}}e-{{\log }_{d}}f={{\log }_{d}}\left( \dfrac{e}{f} \right)$. Let us use this result in equation (4).
$\Rightarrow {{\log }_{3}}\left( \dfrac{{{\log }_{3}}a}{{{\log }_{3}}\dfrac{1}{b}} \right)={{\log }_{3}}3$ ---(5).
We know that if ${{\log }_{a}}d={{\log }_{a}}e$, then we get $d=e$. Let us use this result in equation (5).
$\Rightarrow \dfrac{{{\log }_{3}}a}{{{\log }_{3}}\dfrac{1}{b}}=3$ ---(6).
We know that $\dfrac{{{\log }_{d}}e}{{{\log }_{d}}f}={{\log }_{f}}e$. Let us use this result in equation (6).
$\Rightarrow {{\log }_{\dfrac{1}{b}}}a=3$ ---(7).
We know that if ${{\log }_{d}}e=f$, then $e={{d}^{f}}$. Let us use this result in equation (7).
$\Rightarrow a={{\left( \dfrac{1}{b} \right)}^{3}}$.
$\Rightarrow a=\dfrac{1}{{{b}^{3}}}$.
$\Rightarrow a{{b}^{3}}=1$.
So, we have found the value of $a{{b}^{3}}$ as 1.

So, the correct answer is “Option c”.

Note: We can see that the given problem has a huge amount of calculation so, we need to perform each step carefully in order to avoid confusion and calculation mistakes. We can also solve this problem as shown below:
So, we have ${{\log }_{3}}\left( {{\log }_{3}}a \right)+{{\log }_{\dfrac{1}{3}}}\left( {{\log }_{\dfrac{1}{3}}}b \right)=1$.
We know that ${{\log }_{p}}p=1$.
$\Rightarrow {{\log }_{3}}\left( {{\log }_{3}}a \right)+{{\log }_{\dfrac{1}{3}}}\left( {{\log }_{\dfrac{1}{3}}}b \right)={{\log }_{3}}3$.
We know that ${{\log }_{\dfrac{1}{p}}}q=-{{\log }_{p}}q$.
$\Rightarrow {{\log }_{3}}\left( {{\log }_{3}}a \right)-{{\log }_{3}}\left( {{\log }_{\dfrac{1}{3}}}b \right)={{\log }_{3}}3$.
We know that ${{\log }_{d}}e-{{\log }_{d}}f={{\log }_{d}}\left( \dfrac{e}{f} \right)$.
$\Rightarrow {{\log }_{3}}\left( \dfrac{{{\log }_{3}}a}{{{\log }_{\dfrac{1}{3}}}b} \right)={{\log }_{3}}3$.
We know that if ${{\log }_{a}}d={{\log }_{a}}e$, then we get $d=e$.
$\Rightarrow \dfrac{{{\log }_{3}}a}{{{\log }_{\dfrac{1}{3}}}b}=3$.
$\Rightarrow {{\log }_{3}}a=3{{\log }_{\dfrac{1}{3}}}b$.
We know that $n{{\log }_{a}}d={{\log }_{a}}{{d}^{n}}$.
$\Rightarrow {{\log }_{3}}a={{\log }_{\dfrac{1}{3}}}{{b}^{3}}$.
We know that ${{\log }_{\dfrac{1}{p}}}q=-{{\log }_{p}}q$.
$\Rightarrow {{\log }_{3}}a=-{{\log }_{3}}{{b}^{3}}$.
We know that $-{{\log }_{p}}q={{\log }_{p}}\left( \dfrac{1}{q} \right)$.
$\Rightarrow {{\log }_{3}}a={{\log }_{3}}\left( \dfrac{1}{{{b}^{3}}} \right)$.
We know that if ${{\log }_{a}}d={{\log }_{a}}e$, then we get $d=e$.
$\Rightarrow a=\dfrac{1}{{{b}^{3}}}$.
$\Rightarrow a{{b}^{3}}=1$.