Question

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

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).