Question

If we are given an expression as \[{{\log }_{2}}\left( {{\log }_{8}}x \right)={{\log }_{8}}\left( {{\log }_{2}}x \right)\], then find the value of \[{{\left( {{\log }_{2}}x \right)}^{2}}\].

Answer 1

Hint: For solving the logarithmic equation given in the above question, we must make the base of the logarithmic terms on both sides the same. For this, we have to write $8={{2}^{3}}$ in the given equation and use the logarithmic properties ${{\log }_{{{a}^{b}}}}x=\dfrac{1}{b}{{\log }_{a}}x$ and $k{{\log }_{a}}x={{\log }_{a}}{{x}^{k}}$ to get the base two on both the sides. Finally by comparing both sides of the obtained equation, we will get the required value of \[{{\left( {{\log }_{2}}x \right)}^{2}}\].

Complete step by step solution:
The equation in the above question is written as
$\Rightarrow {{\log }_{2}}\left( {{\log }_{8}}x \right)={{\log }_{8}}\left( {{\log }_{2}}x \right)$
We know that eight is equal to the cube of two. Therefore we can substitute $8={{2}^{3}}$ the above equation to get
$\Rightarrow {{\log }_{2}}\left( {{\log }_{{{2}^{3}}}}x \right)={{\log }_{{{2}^{3}}}}\left( {{\log }_{2}}x \right)$
Now, from the properties of the logarithm function we know that ${{\log }_{{{a}^{b}}}}x=\dfrac{1}{b}{{\log }_{a}}x$. Therefore, we can write the above equation as
$\Rightarrow {{\log }_{2}}\left( \dfrac{1}{3}{{\log }_{2}}x \right)=\dfrac{1}{3}{{\log }_{2}}\left( {{\log }_{2}}x \right)$
From the properties of the logarithm function, we also know that $k{{\log }_{a}}x={{\log }_{a}}{{x}^{k}}$. On applying this property on the RHS of the above equation, we will get
$\Rightarrow {{\log }_{2}}\left( \dfrac{1}{3}{{\log }_{2}}x \right)={{\log }_{2}}{{\left( {{\log }_{2}}x \right)}^{\dfrac{1}{3}}}$
Since the bases on both the sides of the above equation are equal, we can equate the respective arguments to get
$\Rightarrow \dfrac{1}{3}{{\log }_{2}}x={{\left( {{\log }_{2}}x \right)}^{\dfrac{1}{3}}}$
Taking cube on both the sides, we get
\[\begin{align}
  & \Rightarrow {{\left( \dfrac{1}{3}{{\log }_{2}}x \right)}^{3}}={{\left[ {{\left( {{\log }_{2}}x \right)}^{\dfrac{1}{3}}} \right]}^{3}} \\
 & \Rightarrow \dfrac{1}{{{3}^{3}}}{{\left( {{\log }_{2}}x \right)}^{3}}={{\left( {{\log }_{2}}x \right)}^{\dfrac{1}{3}\times 3}} \\
 & \Rightarrow \dfrac{1}{27}{{\left( {{\log }_{2}}x \right)}^{3}}={{\log }_{2}}x \\
\end{align}\]
Subtracting \[{{\log }_{2}}x\] from both the sides, we get
$\begin{align}
  & \Rightarrow \dfrac{1}{27}{{\left( {{\log }_{2}}x \right)}^{3}}-{{\log }_{2}}x={{\log }_{2}}x-{{\log }_{2}}x \\
 & \Rightarrow \dfrac{1}{27}{{\left( {{\log }_{2}}x \right)}^{3}}-{{\log }_{2}}x=0 \\
\end{align}$
Now, we can take ${{\log }_{2}}x$ common on the LHS to get
\[\Rightarrow {{\log }_{2}}x\left[ \dfrac{1}{27}{{\left( {{\log }_{2}}x \right)}^{2}}-1 \right]=0\]
Using the zero product rule we get
$\Rightarrow {{\log }_{2}}x=0.......\left( i \right)$
And
\[\Rightarrow \dfrac{1}{27}{{\left( {{\log }_{2}}x \right)}^{2}}-1=0........\left( ii \right)\]
Taking squares on both the sides of the equation (i) we get
$\begin{align}
  & \Rightarrow {{\left( {{\log }_{2}}x \right)}^{2}}={{0}^{2}} \\
 & \Rightarrow {{\left( {{\log }_{2}}x \right)}^{2}}=0 \\
\end{align}$
Adding $1$ on both sides of the equation (ii) we get
\[\begin{align}
  & \Rightarrow \dfrac{1}{27}{{\left( {{\log }_{2}}x \right)}^{2}}-1+1=0+1 \\
 & \Rightarrow \dfrac{1}{27}{{\left( {{\log }_{2}}x \right)}^{2}}=1 \\
\end{align}\]
Multiplying $27$ both sides, we get
\[\begin{align}
  & \Rightarrow \dfrac{1}{27}{{\left( {{\log }_{2}}x \right)}^{2}}\times 27=1\times 27 \\
 & \Rightarrow {{\left( {{\log }_{2}}x \right)}^{2}}=27 \\
\end{align}\]
Hence, we have got two values of \[{{\left( {{\log }_{2}}x \right)}^{2}}\] which are respectively $0$ and $27$.

Note: For solving these types of questions, we must be familiar with the properties of the logarithmic function. Do not forget to consider the zero value of \[{{\left( {{\log }_{2}}x \right)}^{2}}\], which is obtained in the above solution. Since zero is obtainable from the logarithm function for $x=1$, we cannot reject it.