QUESTION

# The value of $\log {}_{\dfrac{1}{2}}4$ is:a.$-2$b.$0$c.$\dfrac{1}{2}$d.$2$

Hint: Here, we have to make 4 in terms of $\dfrac{1}{2}$ by taking $4={{2}^{2}}$ where ${{2}^{2}}=\dfrac{1}{{{2}^{-2}}}$.Here, we have to apply the logarithmic properties:

\begin{align} & \log \left( {{a}^{x}} \right)=x\log a \\ & \log {}_{a}a=1 \\ & \log \left( \dfrac{a}{b} \right)=\log a-\log b \\ \end{align}

Here, we want to find the value of $\log {}_{\dfrac{1}{2}}4$.
We know that,
$4={{2}^{2}}$

It can also be written as:
\begin{align} & 4=\dfrac{1}{{{2}^{-2}}} \\ & 4={{\left( \dfrac{1}{2} \right)}^{-2}} \\ \end{align}
Here, we can write the given logarithmic function in the form:
${{\log }_{\dfrac{1}{2}}}4={{\log }_{\dfrac{1}{2}}}{{\left( \dfrac{1}{2} \right)}^{-2}}$

By the logarithmic property we know that:
$\log \left( {{a}^{x}} \right)=x\log a$
Therefore, by applying this property we can write:
${{\log }_{\dfrac{1}{2}}}4=-2{{\log }_{\dfrac{1}{2}}}\dfrac{1}{2}$ ….. (1)
Hence, by the logarithmic property we have:
$\log {}_{a}a=1$
Therefore, by applying this property we can write:
$\log {}_{\dfrac{1}{2}}\dfrac{1}{2}=1$
Hence, our equation (1) becomes:
\begin{align} & {{\log }_{\dfrac{1}{2}}}4=-2\times 1 \\ & {{\log }_{\dfrac{1}{2}}}4=-2 \\ \end{align}
Therefore, we can say that the value of ${{\log }_{\dfrac{1}{2}}}4=-2$.

OR

Here, you can also find the value by another method.
Consider the property:
$\log {}_{x}a=\dfrac{\log x}{\log a}$
By applying this property we get:
${{\log }_{\dfrac{1}{2}}}4=\dfrac{\log 4}{\log \dfrac{1}{2}}$

Now we can write:
$4={{2}^{2}}$
Therefore, our equation becomes:
${{\log }_{\dfrac{1}{2}}}4=\dfrac{\log {{2}^{2}}}{\log \dfrac{1}{2}}$ ….. (2)
Hence by the property we know that,
$\log \left( {{a}^{x}} \right)=x\log a$
So we can write:
$\log {{2}^{2}}=2\log 2$

By substituting the above value in equation (2) we obtain:
${{\log }_{\dfrac{1}{2}}}4=\dfrac{2\log 2}{\log \dfrac{1}{2}}$ …. (3)
Now, we also have a property that:
$\log \left( \dfrac{a}{b} \right)=\log a-\log b$
Therefore, we can write:
$\log \dfrac{1}{2}=\log 1-\log 2$
We also know that the value of $\log 1=0$

Hence, we will get:
\begin{align} & \log \dfrac{1}{2}=0-\log 2 \\ & \log \dfrac{1}{2}=-\log 2 \\ \end{align}
Hence, by substituting this value in equation (3) we get:
${{\log }_{\dfrac{1}{2}}}4=\dfrac{2\log 2}{-\log 2}$
Now, by cancellation we obtain:
${{\log }_{\dfrac{1}{2}}}4=-2$

Hence, the correct answer for this question is option (a).

Note: Here, the first method is a much more time consuming method, if you know how to convert 4 in terms of $\dfrac{1}{2}$. Then in two steps you can reach the answer.