Question

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

Answer 1

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.