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(a) \[{{\log }_{2}}3\]

(b) \[{{\log }_{\dfrac{1}{2}}}5\]

(c) Both are equal

(d) Canâ€™t say

Answer
Verified

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, \[-2**Therefore, 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.

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, \[-2

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.

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