
How do you find the inverse of $y={{\log }_{\dfrac{1}{2}}}x$ ?
Answer
542.4k+ views
Hint: To find the inverse of the given logarithmic function, we are going to replace y by x and x by y then we are going to use the following property of logarithm which is equal to:
${{\log }_{a}}b=m\Rightarrow b={{a}^{m}}$. And then replace y by ${{f}^{-1}}\left( x \right)$. This ${{f}^{-1}}\left( x \right)$ is the inverse of the logarithmic function.
Complete step by step solution:
In the above problem, we have given the following logarithmic function:
$y={{\log }_{\dfrac{1}{2}}}x$ …………. (1)
Now, we are going to replace y by x and x by y in the above equation we get,
$x={{\log }_{\dfrac{1}{2}}}y$ ………. (2)
Now, we are going to write y in terms of x by using the following logarithm property:
$\begin{align}
& {{\log }_{a}}b=m \\
& \Rightarrow b={{a}^{m}} \\
\end{align}$
Now, comparing the above property with eq. (2) we get,
$b=y,a=\dfrac{1}{2},m=x$
So, eq. (2) will become:
$y={{\left( \dfrac{1}{2} \right)}^{x}}$
Now, we are going to replace y by ${{f}^{-1}}\left( x \right)$ in the above equation and we get,
$\Rightarrow {{f}^{-1}}\left( x \right)={{\left( \dfrac{1}{2} \right)}^{x}}$
From the above solution, we have found the inverse of the given logarithm and it is equal to ${{\left( \dfrac{1}{2} \right)}^{x}}$.
Note: The mistake that could be possible in the above problem is in converting the logarithm form into the base form:
${{\log }_{a}}b=m\Rightarrow b={{a}^{m}}$
The blunder which can be possible in the above problem is that you might write the above equation as follows:
$b={{m}^{a}}$
To avoid such mistake, check the value of b whether it is right or not by substituting this value of b in logarithmic expression i.e. ${{\log }_{a}}b$ then see whether we are getting this value of logarithm as m or not.
${{\log }_{a}}{{m}^{a}}$
Here, you will catch your mistake because we cannot use the property of logarithm which states that:
${{\log }_{a}}{{a}^{m}}=m$
This means that the value of b which we have solved is incorrect.
${{\log }_{a}}b=m\Rightarrow b={{a}^{m}}$. And then replace y by ${{f}^{-1}}\left( x \right)$. This ${{f}^{-1}}\left( x \right)$ is the inverse of the logarithmic function.
Complete step by step solution:
In the above problem, we have given the following logarithmic function:
$y={{\log }_{\dfrac{1}{2}}}x$ …………. (1)
Now, we are going to replace y by x and x by y in the above equation we get,
$x={{\log }_{\dfrac{1}{2}}}y$ ………. (2)
Now, we are going to write y in terms of x by using the following logarithm property:
$\begin{align}
& {{\log }_{a}}b=m \\
& \Rightarrow b={{a}^{m}} \\
\end{align}$
Now, comparing the above property with eq. (2) we get,
$b=y,a=\dfrac{1}{2},m=x$
So, eq. (2) will become:
$y={{\left( \dfrac{1}{2} \right)}^{x}}$
Now, we are going to replace y by ${{f}^{-1}}\left( x \right)$ in the above equation and we get,
$\Rightarrow {{f}^{-1}}\left( x \right)={{\left( \dfrac{1}{2} \right)}^{x}}$
From the above solution, we have found the inverse of the given logarithm and it is equal to ${{\left( \dfrac{1}{2} \right)}^{x}}$.
Note: The mistake that could be possible in the above problem is in converting the logarithm form into the base form:
${{\log }_{a}}b=m\Rightarrow b={{a}^{m}}$
The blunder which can be possible in the above problem is that you might write the above equation as follows:
$b={{m}^{a}}$
To avoid such mistake, check the value of b whether it is right or not by substituting this value of b in logarithmic expression i.e. ${{\log }_{a}}b$ then see whether we are getting this value of logarithm as m or not.
${{\log }_{a}}{{m}^{a}}$
Here, you will catch your mistake because we cannot use the property of logarithm which states that:
${{\log }_{a}}{{a}^{m}}=m$
This means that the value of b which we have solved is incorrect.
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