Question

The value of \[\sqrt {\left( {\log _{0.5}^24} \right)} \] is

Answer 1

Hint:
Here, we are required to find the value of the given logarithmic expression. We will use the properties of logarithm to further simplify the given expression. Logarithms are basically the inverse of exponents. After applying the properties of logarithm and finding the square root of the resultant number, we will get the required value of the given expression.

Formula Used:
Some properties of logarithms we will use in this question are:
1) \[{\log _b}a = \dfrac{{\log a}}{{\log b}}\]
2) \[\log {b^a} = a\log b\]
3) \[\log \left( {\dfrac{b}{a}} \right) = - \log \left( {\dfrac{a}{b}} \right)\]
4) \[{\log _a}a = \dfrac{{\log a}}{{\log a}} = 1\]

Complete step by step solution:
We have to find the value of \[\sqrt {\left( {\log _{0.5}^24} \right)} \].
Now, this can be written as \[\sqrt {{{\left( {{{\log }_{0.5}}4} \right)}^2}} \]
By using the property of logarithms \[{\log _b}a = \dfrac{{\log a}}{{\log b}}\], we get
\[\sqrt {{{\left( {{{\log }_{0.5}}4} \right)}^2}} = \sqrt {{{\left( {\dfrac{{\log 4}}{{\log 0.5}}} \right)}^2}} \]
Substituting \[4 = {2^2}\] and \[0.5 = \dfrac{5}{{10}} = \dfrac{1}{2} = {2^{ - 1}}\] in the above equation, we get
\[ \Rightarrow \sqrt {{{\left( {\dfrac{{\log 4}}{{\log 0.5}}} \right)}^2}} = \sqrt {{{\left( {\dfrac{{\log {2^2}}}{{\log {2^{ - 1}}}}} \right)}^2}} \]
Again, using the properties of logarithms \[\log {b^a} = a\log b\] and \[\log \left( {\dfrac{b}{a}} \right) = - \log \left( {\dfrac{a}{b}} \right)\], we get
 \[ \Rightarrow \sqrt {{{\left( {\dfrac{{\log {2^2}}}{{\log {2^{ - 1}}}}} \right)}^2}} = \sqrt {{{\left( {\dfrac{{2\log 2}}{{ - \log 2}}} \right)}^2}} \]
Using the property \[{\log _a}a = \dfrac{{\log a}}{{\log a}} = 1\], we get
\[\begin{array}{l} \Rightarrow \sqrt {{{\left( {\dfrac{{2\log 2}}{{ - \log 2}}} \right)}^2}} = \sqrt {{{\left( { - 2} \right)}^2}} \\ \Rightarrow \sqrt {{{\left( {\dfrac{{2\log 2}}{{ - \log 2}}} \right)}^2}} = \sqrt 4 = 2\end{array}\]

Hence, the value of \[\sqrt {\left( {\log _{0.5}^24} \right)} \] is equal to 2.

Note:
An alternate way to solve this question is:
Taking the square outside the bracket
\[\sqrt {\left( {\log _{0.5}^24} \right)} = \sqrt {{{\left( {{{\log }_{0.5}}4} \right)}^2}} \]
\[ \Rightarrow \sqrt {\left( {\log _{0.5}^24} \right)} = \left| {{{\log }_{0.5}}4} \right|\]
Rewriting the above equation, we get
\[ \Rightarrow \left| {{{\log }_{0.5}}4} \right| = \left| {{{\log }_{{2^{ - 1}}}}{2^2}} \right|\]
Using \[\log {b^a} = a\log b\] and \[\log \left( {\dfrac{b}{a}} \right) = - \log \left( {\dfrac{a}{b}} \right)\], we get
\[ \Rightarrow \left| {{{\log }_{{2^{ - 1}}}}{2^2}} \right| = \left| { - 2{{\log }_2}2} \right|\]
Now, using \[{\log _a}a = \dfrac{{\log a}}{{\log a}} = 1\], we get
\[ \Rightarrow \left| { - 2{{\log }_2}2} \right| = \left| { - 2} \right| = 2\]
Here, we have used the modulus sign because we were required to find the value of \[\sqrt {\left( {\log _{0.5}^24} \right)} \] and the square root can never be negative. Hence, we have used the modulus because it always gives a non-negative value of the number present inside it.
There are chances that we will leave the answer as \[ - 2\] instead of 2 and hence, making the answer completely wrong. Therefore, it is really important to know both the ways by which this question can be solved.