# Evaluate the expression \[{{\log }_{2}}xy=5,{{\log }_{\dfrac{1}{2}}}\left( \dfrac{x}{y} \right)=1\]

Last updated date: 24th Mar 2023

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Answer

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Hint: Use basic identity of logarithm given by;

If ${{a}^{x}}=N\text{ then }{{\log }_{a}}N=x$

We have equations/expression given in the problem as

\[{{\log }_{2}}xy=5.............\left( 1 \right)\]

And

\[{{\log }_{\dfrac{1}{2}}}\left( \dfrac{x}{y} \right)=1....................\left( 2 \right)\]

As, we know that if ${{a}^{x}}=N$ then we can take log to both sides as base of $a$

And we get:

${{a}^{x}}=N$

Taking $\log $ on both sides

${{\log }_{a}}{{a}^{x}}={{\log }_{a}}N$

As we know that ${{\log }_{c}}{{m}^{n}}=n{{\log }_{c}}m$ ;

Using this property we can write the above equation as;

$x{{\log }_{a}}a={{\log }_{a}}N$

As we know ${{\log }_{m}}m=1$ , we can rewrite the above relation as;

${{\log }_{a}}N=x$

Therefore, if we have ${{a}^{x}}=N$

Then ${{\log }_{a}}N=x................\left( 3 \right)$

Using the above property of logarithm we can write equation $\left( 1 \right)$ as

${{\log }_{2}}xy=5$

$xy={{2}^{5}}.................\left( 4 \right)$

Similarly, using the equation $\left( 3 \right)$ , we can write equation $\left( 2 \right)$ as

$\begin{align}

& {{\log }_{\dfrac{1}{2}}}\left( \dfrac{x}{y} \right)=1 \\

& \dfrac{x}{y}={{\left( \dfrac{1}{2} \right)}^{1}}=\left( \dfrac{1}{2} \right)..............\left( 5 \right) \\

\end{align}$

Now, we need to find $x\text{ and y}$ ; For that we can multiply equation $\left( 4 \right)\text{ and }\left( 5 \right)$ in following way;

$\begin{align}

& xy\times \dfrac{x}{y}={{2}^{5}}\times \dfrac{1}{2} \\

& {{x}^{2}}=\dfrac{32}{2}=16 \\

& {{x}^{2}}=16 \\

& x=\pm 4 \\

\end{align}$

To get value of $y$ , we can divide equation $\left( 4 \right)\And \left( 5 \right)$

$\begin{align}

& \dfrac{xy}{\left( \dfrac{x}{y} \right)}=\dfrac{{{2}^{5}}}{\left( \dfrac{1}{2} \right)} \\

& xy\times \dfrac{y}{x}=32\times 2 \\

& {{y}^{2}}=64 \\

& y=\pm 8 \\

\end{align}$

Hence, we have $x=\pm 4\text{ and }y=\pm 8$ .

Now, here we need to select $\left( x,y \right)$ pairs which will satisfy the equation$\left( 5 \right)\And \left( 4 \right)$.

Now, we have four pairs as

$\begin{align}

& x=4,y=8 \\

& x=-4,y=-8 \\

& x=4,y=-8 \\

& x=-4,y=8 \\

\end{align}$

We can put pairs to equation $\left( 4 \right)\And \left( 5 \right)$for verification

Case 1: $x=4,y=8$

For equation $\left( 4 \right)\text{ }xy=32$

$LHS=4\times 8=32=RHS$

For equation \[\left( 5 \right)\text{ }\dfrac{x}{y}=\dfrac{1}{2}\]

\[LHS=\dfrac{4}{8}=\dfrac{1}{2}=RHS\]

Hence $\left( 4,8 \right)$ is the solution of the given equations.

Case 2: $x=-4,y=8$

For equation $\left( 4 \right)$ $xy=32$

$LHS=-4\times -8=32=RHS$

For equation $\left( 5 \right)\text{ }\dfrac{x}{y}=\dfrac{1}{2}$

\[LHS=\dfrac{-4}{-8}=\dfrac{1}{2}=RHS\]

Hence, $\left( -4,-8 \right)$ pair is also a solution of the given equations.

Case 3: $x=-4,y=8$

For equation $\left( 4 \right)\text{ }xy=32$

$LHS=-4\times 8=-32\ne RHS$

It will not satisfy the equation $\left( 5 \right)$ $\dfrac{x}{y}=\dfrac{1}{2}$ as well.

Hence, $\left( -4,8 \right)$ pair is not a solution to the given equation.

Case 4: $x=4,y=-8$

For equation $\left( 4 \right)\text{ }xy=32$

$4\times \left( -8 \right)=-32\ne RHS$

For equation $\left( 5 \right)\dfrac{x}{y}=\dfrac{1}{2}$

$LHS=\dfrac{4}{-8}=-\dfrac{1}{2}\ne RHS$

Hence,$\left( 4,-8 \right)$ is not a solution of the given equation.

Note: We can get answers by putting values of $x=\pm 4$ in any of the equation $\left( 3 \right)\And \left( 4 \right)$ which will minimize our confusion related to $\left( -4,8 \right)or\left( 4,-8 \right)$ as explained in solution. One can also skip the question by just seeing the solution by just seeing the given function $\left( {{\log }_{2}}xy=5\text{ }\!\!\And\!\!\text{ }{{\log }_{\dfrac{1}{2}}}\dfrac{x}{y}=1 \right)$ as we cannot put negative values in logarithm $m$ function. Domain of $\log x$ is ${{R}^{+}}$ (positive real numbers).

One can go wrong by getting confused with formula if ${{a}^{x}}=N$ then ${{\log }_{a}}N=x$ . He/she may apply if ${{a}^{x}}=N$then ${{\log }_{N}}a=x$(general confusion with basic definition of logarithm function).

If ${{a}^{x}}=N\text{ then }{{\log }_{a}}N=x$

We have equations/expression given in the problem as

\[{{\log }_{2}}xy=5.............\left( 1 \right)\]

And

\[{{\log }_{\dfrac{1}{2}}}\left( \dfrac{x}{y} \right)=1....................\left( 2 \right)\]

As, we know that if ${{a}^{x}}=N$ then we can take log to both sides as base of $a$

And we get:

${{a}^{x}}=N$

Taking $\log $ on both sides

${{\log }_{a}}{{a}^{x}}={{\log }_{a}}N$

As we know that ${{\log }_{c}}{{m}^{n}}=n{{\log }_{c}}m$ ;

Using this property we can write the above equation as;

$x{{\log }_{a}}a={{\log }_{a}}N$

As we know ${{\log }_{m}}m=1$ , we can rewrite the above relation as;

${{\log }_{a}}N=x$

Therefore, if we have ${{a}^{x}}=N$

Then ${{\log }_{a}}N=x................\left( 3 \right)$

Using the above property of logarithm we can write equation $\left( 1 \right)$ as

${{\log }_{2}}xy=5$

$xy={{2}^{5}}.................\left( 4 \right)$

Similarly, using the equation $\left( 3 \right)$ , we can write equation $\left( 2 \right)$ as

$\begin{align}

& {{\log }_{\dfrac{1}{2}}}\left( \dfrac{x}{y} \right)=1 \\

& \dfrac{x}{y}={{\left( \dfrac{1}{2} \right)}^{1}}=\left( \dfrac{1}{2} \right)..............\left( 5 \right) \\

\end{align}$

Now, we need to find $x\text{ and y}$ ; For that we can multiply equation $\left( 4 \right)\text{ and }\left( 5 \right)$ in following way;

$\begin{align}

& xy\times \dfrac{x}{y}={{2}^{5}}\times \dfrac{1}{2} \\

& {{x}^{2}}=\dfrac{32}{2}=16 \\

& {{x}^{2}}=16 \\

& x=\pm 4 \\

\end{align}$

To get value of $y$ , we can divide equation $\left( 4 \right)\And \left( 5 \right)$

$\begin{align}

& \dfrac{xy}{\left( \dfrac{x}{y} \right)}=\dfrac{{{2}^{5}}}{\left( \dfrac{1}{2} \right)} \\

& xy\times \dfrac{y}{x}=32\times 2 \\

& {{y}^{2}}=64 \\

& y=\pm 8 \\

\end{align}$

Hence, we have $x=\pm 4\text{ and }y=\pm 8$ .

Now, here we need to select $\left( x,y \right)$ pairs which will satisfy the equation$\left( 5 \right)\And \left( 4 \right)$.

Now, we have four pairs as

$\begin{align}

& x=4,y=8 \\

& x=-4,y=-8 \\

& x=4,y=-8 \\

& x=-4,y=8 \\

\end{align}$

We can put pairs to equation $\left( 4 \right)\And \left( 5 \right)$for verification

Case 1: $x=4,y=8$

For equation $\left( 4 \right)\text{ }xy=32$

$LHS=4\times 8=32=RHS$

For equation \[\left( 5 \right)\text{ }\dfrac{x}{y}=\dfrac{1}{2}\]

\[LHS=\dfrac{4}{8}=\dfrac{1}{2}=RHS\]

Hence $\left( 4,8 \right)$ is the solution of the given equations.

Case 2: $x=-4,y=8$

For equation $\left( 4 \right)$ $xy=32$

$LHS=-4\times -8=32=RHS$

For equation $\left( 5 \right)\text{ }\dfrac{x}{y}=\dfrac{1}{2}$

\[LHS=\dfrac{-4}{-8}=\dfrac{1}{2}=RHS\]

Hence, $\left( -4,-8 \right)$ pair is also a solution of the given equations.

Case 3: $x=-4,y=8$

For equation $\left( 4 \right)\text{ }xy=32$

$LHS=-4\times 8=-32\ne RHS$

It will not satisfy the equation $\left( 5 \right)$ $\dfrac{x}{y}=\dfrac{1}{2}$ as well.

Hence, $\left( -4,8 \right)$ pair is not a solution to the given equation.

Case 4: $x=4,y=-8$

For equation $\left( 4 \right)\text{ }xy=32$

$4\times \left( -8 \right)=-32\ne RHS$

For equation $\left( 5 \right)\dfrac{x}{y}=\dfrac{1}{2}$

$LHS=\dfrac{4}{-8}=-\dfrac{1}{2}\ne RHS$

Hence,$\left( 4,-8 \right)$ is not a solution of the given equation.

Note: We can get answers by putting values of $x=\pm 4$ in any of the equation $\left( 3 \right)\And \left( 4 \right)$ which will minimize our confusion related to $\left( -4,8 \right)or\left( 4,-8 \right)$ as explained in solution. One can also skip the question by just seeing the solution by just seeing the given function $\left( {{\log }_{2}}xy=5\text{ }\!\!\And\!\!\text{ }{{\log }_{\dfrac{1}{2}}}\dfrac{x}{y}=1 \right)$ as we cannot put negative values in logarithm $m$ function. Domain of $\log x$ is ${{R}^{+}}$ (positive real numbers).

One can go wrong by getting confused with formula if ${{a}^{x}}=N$ then ${{\log }_{a}}N=x$ . He/she may apply if ${{a}^{x}}=N$then ${{\log }_{N}}a=x$(general confusion with basic definition of logarithm function).

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