Question

The value of the expression given as \[{{\left( {{\left( {{\log }_{2}}9 \right)}^{2}} \right)}^{\dfrac{1}{{{\log }_{2}}{{\log }_{2}}9}}}\times {{\left( \sqrt{7} \right)}^{\dfrac{1}{{{\log }_{4}}7}}}\]

Answer 1

Hint: To solve this question we will try to resolve the given expression by using log property. First we will use $\dfrac{1}{{{\log }_{a}}b}={{\log }_{b}}a$ to reduce the power term having $\dfrac{1}{{{\log }_{2}}{{\log }_{2}}9}$ as the power, then we will use the log property $m{{\log }_{n}}=\log {{n}^{m}}$ for clearing any remaining powers of expression. Then, finally we will use ${{a}^{{{\log }_{a}}b}}=b$ to get our result.

Complete step-by-step solution:
We know that, \[\dfrac{1}{{{\log }_{a}}b}={{\log }_{b}}a\]
We have our expression as \[{{\left( {{\left( {{\log }_{2}}9 \right)}^{2}} \right)}^{\dfrac{1}{{{\log }_{2}}{{\log }_{2}}9}}}\times {{\left( \sqrt{7} \right)}^{\dfrac{1}{{{\log }_{4}}7}}}\]
Using the property of log stated above in our expression by using $b={{\log }_{2}}9\text{ and a}=2$ we get:
\[\begin{align}
  & {{\left( {{\left( {{\log }_{2}}9 \right)}^{2}} \right)}^{\dfrac{1}{{{\log }_{2}}{{\log }_{2}}9}}}\times {{\left( \sqrt{7} \right)}^{\dfrac{1}{{{\log }_{4}}7}}} \\
 & {{\left( {{\log }_{2}}9 \right)}^{lo{{g}_{{{\log }_{2}}9}}2}}\times {{\left( \sqrt{7} \right)}^{\dfrac{1}{{{\log }_{4}}7}}} \\
\end{align}\]
Now, changing $\sqrt{7}$ to ${{\left( 7 \right)}^{\dfrac{1}{2}}}$ and using log identity \[\dfrac{1}{{{\log }_{a}}b}={{\log }_{b}}a\] for the term \[{{\log }_{4}}7\] by substituting a = 4 and b = 7 we get:
\[{{\left( {{\left( {{\log }_{2}}9 \right)}^{2}} \right)}^{{{\log }_{{{\log }_{2}}9}}2}}\times {{\left( 7 \right)}^{\dfrac{1}{2}{{\log }_{7}}4}}\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. (i)}\]
Now, let us assume $x={{7}^{\dfrac{1}{2}{{\log }_{7}}4}}$ we will solve x part first:
\[\begin{align}
  & x={{7}^{\dfrac{1}{2}{{\log }_{7}}4}} \\
 & \text{as }{{\text{2}}^{\text{2}}}=4 \\
 & x={{7}^{\dfrac{1}{2}{{\log }_{7}}{{4}^{2}}}} \\
\end{align}\]
Also, we have a log property that $m{{\log }_{n}}=\log {{n}^{m}}$ using this above by substituting $m=\dfrac{1}{2}\text{ and }n={{2}^{2}}$ we get:
\[\begin{align}
  & x={{7}^{{{\log }_{7}}}}^{{{\left( {{2}^{2}} \right)}^{\dfrac{1}{2}}}} \\
 & x={{7}^{{{\log }_{7}}2}} \\
\end{align}\]
Substituting this result in equation (i) we have:
\[{{\left( {{\left( {{\log }_{2}}9 \right)}^{2}} \right)}^{{{\log }_{{{\log }_{2}}9}}2}}\times {{\left( 7 \right)}^{{{\log }_{7}}2}}\]
Now using property $m{{\log }_{n}}=\log {{n}^{m}}$ in above by taking m = 2 and n = 2 for the first term we get:
\[\begin{align}
  & {{\left( {{\log }_{2}}9 \right)}^{{{\log }_{{{\log }_{2}}9}}{{2}^{2}}}}\times {{\left( 7 \right)}^{{{\log }_{7}}2}} \\
 & {{\left( {{\log }_{2}}9 \right)}^{lo{{g}_{{{\log }_{2}}9}}4}}\times {{\left( 7 \right)}^{{{\log }_{7}}2}} \\
\end{align}\]
Again we have property of log which says that ${{a}^{{{\log }_{a}}b}}=b$
Using this in above, substitute $a={{\log }_{2}}9\text{ and b}=4$ we get:
\[\begin{align}
  & {{\left( {{\log }_{2}}9 \right)}^{{{\log }_{{{\log }_{2}}9}}4}}\times {{\left( 7 \right)}^{{{\log }_{7}}2}} \\
 & \Rightarrow 4\times {{\left( 7 \right)}^{{{\log }_{7}}2}} \\
\end{align}\]
Again, using property ${{a}^{{{\log }_{a}}b}}=b$ by substituting a = 7 and b = 2 we have:
\[\Rightarrow 4\times 2=8\]
So, the value of \[{{\left( {{\left( {{\log }_{2}}9 \right)}^{2}} \right)}^{\dfrac{1}{{{\log }_{2}}{{\log }_{2}}9}}}\times {{\left( \sqrt{7} \right)}^{\dfrac{1}{{{\log }_{4}}7}}}\] is 8.

Note: Students can get confuse at the point where we are applying log property given as \[\dfrac{1}{{{\log }_{a}}b}={{\log }_{b}}a\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. (ii)}\]
Here, in this question we have used the above property of log in the term \[{{\left( {{\left( {{\log }_{2}}9 \right)}^{2}} \right)}^{\dfrac{1}{{{\log }_{2}}{{\log }_{2}}9}}}\]
We have only applied to power of ${{\log }_{2}}9$ that is to \[\dfrac{2}{{{\log }_{2}}{{\log }_{2}}9}\]
Now, compare this formula as given in equation (ii) above, we will use \[b={{\log }_{2}}9\text{ and a}=2\]
Then, after applying this formula we get: \[\dfrac{1\times 2}{{{\log }_{2}}{{\log }_{2}}9}=2{{\log }_{{{\log }_{2}}9}}2\]
So, in RHS we have \[{{\log }_{2}}9\] in the base of log2. So this point of confusion is cleared.