Which one of the two is greater? \[{{\log }_{2}}3\]or \[{{\log }_{\dfrac{1}{2}}}5\]
(a) \[{{\log }_{2}}3\]
(b) \[{{\log }_{\dfrac{1}{2}}}5\]
(c) Both are equal
(d) Can’t say

Hint: Convert \[{{\log }_{a}}x=y\]to \[x={{a}^{y}}\]and find the appropriate values of \[{{a}^{y}}\].

 Here, we have to find that which is greater among \[{{\log }_{2}}3\]or \[{{\log }_{\dfrac{1}{2}}}5\].
Taking \[{{\log }_{2}}3=A\]
We know that, when \[{{\log }_{b}}a=n\]
Then, \[a={{b}^{n}}....\left( i \right)\]
Similarly, \[3={{2}^{A}}...\left( ii \right)\]
Now, we know that \[{{\left( 2 \right)}^{1}}=2\]and \[{{\left( 2 \right)}^{2}}=4\].
Also, \[{{2}^{1}}<3<{{2}^{2}}\]
From equation \[\left( ii \right)\], \[{{2}^{1}}<{{2}^{A}}<{{2}^{2}}\]
Hence, \[1Therefore, we get the approximate value of \[A={{\log }_{2}}3\]between \[1\] and \[2\].
Taking \[{{\log }_{\dfrac{1}{2}}}5=B\]
From equation \[\left( i \right)\],
\[\Rightarrow 5={{\left( \dfrac{1}{2} \right)}^{B}}\]
We know that \[\dfrac{1}{a}={{a}^{-1}}\]
Hence, we get \[5={{2}^{-B}}....\left( iii \right)\]
Now, we know that
\[{{2}^{2}}=4\]
And \[{{2}^{3}}=8\]
Also, \[{{2}^{2}}<5<{{2}^{3}}\]
As, \[-\left( -a \right)=a\]
We can also write it as \[{{2}^{-\left( -2 \right)}}<5<{{2}^{-\left( -3 \right)}}\]
From equation \[\left( iii \right)\], \[{{2}^{-\left( -2 \right)}}<{{2}^{-B}}<{{2}^{-\left( -3 \right)}}\]
Hence, \[-2Therefore, we get the approximate value of \[B={{\log }_{\dfrac{1}{2}}}5\]between \[-2\]and \[-3\].
Clearly, as \[A\]is positive and \[B\]is negative.
\[A>B\]
Or, \[{{\log }_{2}}3>{{\log }_{\dfrac{1}{2}}}5\]
Therefore, option (a) is correct

Note: Here, some students may think that as \[\dfrac{1}{2}\]is smaller as compared to \[2\], so it will require greater power to become \[5\]as compared to power required by\[2\] to become \[3\]. But in this process, they miss the negative sign in the power that will also be required to convert \[\dfrac{1}{2}\]to \[5\]and get the wrong result.