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

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:

& {{\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{aligned}$$ Now, changing $\sqrt{7}$ to ${{\left( 7 \right)}^{\dfrac{1}{2}}}$ and using log identity $$\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{aligned} & 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{aligned}$$ Also, we have a log property that $m{{\log }_{n}}=\log {{n}^{m}}$ using this above by substituting $m=\dfrac{1}{2}\text{ and }n={{2}^{2}}$ we get: $$\begin{aligned} & x={{7}^{{{\log }_{7}}}}^{{{\left( {{2}^{2}} \right)}^{\dfrac{1}{2}}}} \\\ & x={{7}^{{{\log }_{7}}2}} \\\ \end{aligned}$$ Substituting this result in equation (i) we have: $${{\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{aligned} & {{\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{aligned}$$ Again we have property of log which says that ${{a}^{{{\log }_{a}}b}}=b$ Using this in above, substitute $a={{\log }_{2}}9\text{ and b}=4$ we get: $$\begin{aligned} & {{\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{aligned}$$ Again, using property ${{a}^{{{\log }_{a}}b}}=b$ by substituting a = 7 and b = 2 we have: $$\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.