Question

Find the value of \[8\left( {{\log }_{\sqrt{x}}}\sqrt{x\sqrt{x\sqrt{x\sqrt{x}}}} \right)\] ?

Answer 1

Hint: In the given question, we have been asked to find the value of a given expression. In order to solve the question, first we need to apply the property of logarithm i.e. \[{{\log }_{a}}{{a}^{n}}=n\] later simplifying the expression, we will need to use the laws of exponent of power i.e. \[{{a}^{m}}\times {{a}^{n}}={{a}^{m+n}}\] and \[{{\left( {{a}^{m}} \right)}^{n}}={{a}^{m\times n}}\] . By using these properties we will simplify the expression till we get the value. In this way we will get the required answer.

Complete step-by-step answer:
We have given that:
 \[8\left( {{\log }_{\sqrt{x}}}\sqrt{x\sqrt{x\sqrt{x\sqrt{x}}}} \right)\]
Using the definition of logarithm which states that,
 \[{{\log }_{a}}{{a}^{n}}=n\]
Applying this property of logarithmic function, we will obtain
As we know that \[\sqrt{x}={{x}^{\dfrac{1}{2}}}\]
 \[\Rightarrow 8\dfrac{{{\log }_{x}}\left( \sqrt{x\sqrt{x\sqrt{{{x}^{1}}\cdot {{x}^{\dfrac{1}{2}}}}}} \right)}{{{\log }_{x}}\left( {{x}^{\dfrac{1}{2}}} \right)}\]
Using the laws of exponent and power which states that \[{{a}^{m}}\times {{a}^{n}}={{a}^{m+n}}\]
Applying this property in the above expression, we will get
 \[\Rightarrow 8\dfrac{{{\log }_{x}}\left( \sqrt{x\sqrt{x\sqrt{{{x}^{\dfrac{3}{2}}}}}} \right)}{\dfrac{1}{2}}\]
Simplifying the above, we get
 \[\Rightarrow 8\dfrac{{{\log }_{x}}\left( \sqrt{x\sqrt{x\cdot {{\left( {{x}^{\dfrac{3}{2}}} \right)}^{\dfrac{1}{2}}}}} \right)}{\dfrac{1}{2}}\]
Using the power property of exponent which states that \[{{\left( {{a}^{m}} \right)}^{n}}={{a}^{m\times n}}\]
Thus, we get
 \[\Rightarrow 8\dfrac{{{\log }_{x}}\left( \sqrt{x\sqrt{x\cdot {{x}^{\dfrac{3}{4}}}}} \right)}{\dfrac{1}{2}}\]
Again,
Using the laws of exponent and power which states that \[{{a}^{m}}\times {{a}^{n}}={{a}^{m+n}}\]
Applying this property in the above expression, we will get
 \[\Rightarrow 8\dfrac{{{\log }_{x}}\left( \sqrt{x\sqrt{{{x}^{\dfrac{7}{4}}}}} \right)}{\dfrac{1}{2}}\]
Simplifying the above, we get
 \[\Rightarrow 8\dfrac{{{\log }_{x}}\left( \sqrt{x{{\left( {{x}^{\dfrac{7}{4}}} \right)}^{\dfrac{1}{2}}}} \right)}{\dfrac{1}{2}}=16{{\log }_{x}}\left( \sqrt{x\cdot {{x}^{\dfrac{7}{8}}}} \right)\]
Similarly using the laws of exponent and power, we get
 \[\Rightarrow 16{{\log }_{x}}\left( \sqrt{{{x}^{1}}\cdot {{x}^{\dfrac{7}{8}}}} \right)=16{{\log }_{x}}\left( \sqrt{{{x}^{1+\dfrac{7}{8}}}} \right)=16{{\log }_{x}}\left( \sqrt{{{x}^{\dfrac{15}{8}}}} \right)=16{{\log }_{x}}\cdot {{\left( {{x}^{\dfrac{15}{8}}} \right)}^{\dfrac{1}{2}}}\]
Simplifying the above expression, we will get
 \[\Rightarrow 16{{\log }_{x}}\cdot {{\left( {{x}^{\dfrac{15}{8}}} \right)}^{\dfrac{1}{2}}}=16{{\log }_{x}}{{x}^{\dfrac{15}{16}}}\]
Using the definition of logarithm which states that,
 \[{{\log }_{a}}{{a}^{n}}=n\]
Applying this property of logarithmic function, we will obtain
We have, \[{{\log }_{x}}{{x}^{\dfrac{15}{16}}}\] which is equal to \[\dfrac{15}{16}\]
Thus,
 \[\Rightarrow 16{{\log }_{x}}{{x}^{\dfrac{15}{16}}}=16\times \dfrac{15}{16}=15\]
Therefore,
 \[\Rightarrow 8\left( {{\log }_{\sqrt{x}}}\sqrt{x\sqrt{x\sqrt{x\sqrt{x}}}} \right)=15\]
Hence, this is the required answer.

Note: While solving these types of questions, it is necessary that students must need to know about the basic properties of logarithmic function. They should also be very aware about the laws of exponent and power which helps while solving the powers and exponents and the given question. You should be very careful while doing the calculation to avoid making any error.