How do you find the inverse of $y={{\left( \dfrac{1}{2} \right)}^{x}}?$
Answer
608.4k+ views
Hint: The given equation we have to make its inverse function.
Inverse function means we have to reverse that original function or equation. For example, if $f$ is from $a$ to $b$ them its inverse will be ${{f}^{-1}}$ from $b$ to $a$.
i.e. $f(x)=b$ and ${{f}^{-1}}\left( b \right)=a$
First write the equation as it is given in the question. Then interchange the variable on both sides and take the logarithm on both sides.
Complete step-by-step answer:
The equation given in the question is as,
$y={{\left( \dfrac{1}{2} \right)}^{x}}...(i)$
Now, let’s interchange the variable so that we get,
$x={{\left( \dfrac{1}{2} \right)}^{y}}...(ii)$
Then we have to solve for $y.$
Now, take logarithm on both side of equation $(i)$ with base $\dfrac{1}{2}$
${{\log }_{\dfrac{1}{2}}}x={{\log }_{\dfrac{1}{2}}}{{\left( \dfrac{1}{2} \right)}^{y}}$
${{\log }_{\dfrac{1}{2}}}=y$
We can write ${{\log }_{\dfrac{1}{2}}}$ as ${{\log }_{2}}-1$
But according to logarithm rules.
${{\log }_{a}}-1=-{{\log }_{a}}$
$y=-{{\log }_{2}}x$
So, the graph of $y={{\left( \dfrac{1}{2} \right)}^{x}}$ and the inverse function of it is $y=-{{\log }_{2}}x$
Additional Information:
Let, $f$ is a function of $1-1$ along with domain $A$ and range $B.$
Then, the inverse function of above function will be as follows with domain \[B\] and range $A,$
The inverse of $f$ is denoted as ${{f}^{-1}}$
${{f}^{-1}}\left( y \right)=x$
Then $f(x)=y$ for any $y\in B$
Remember that inverse functions are only applicable for $1-1$ functions.
For example, if $h\left( -5 \right)=8$ then find ${{h}^{-1}}\left( 8 \right)=?$
Given, $h=\left( -5 \right)=8$
Let, $h\left( x \right)$ is the graph where $\left( -5,8 \right)$ points are located.
Now, inverse the graph $h(x)$ i.e. ${{h}^{-1}}\left( x \right)$ then the points on it will also get inverse i.e. $\left( 8,-5 \right)$on the graph.
So, when $h(x)=\left( -5,8 \right)$ after inversing ${{h}^{-1}}\left( x \right)=\left( 8,-5 \right)$
From above, we can say that,
If $h\left( -5 \right)=8$
Then, ${{h}^{-1}}\left( 8 \right)=-5$
Note:
Here we are taking the logarithm rule for simplifying the equation. We also interchange the variables which are not necessary. We can also solve it without interchanging the variables.
Then the solution will becomes,
After taking logarithm on both sides of the equation $(i)$.
${{\log }_{\dfrac{1}{2}}}y={{\log }_{\dfrac{1}{2}}}{{\left( \dfrac{1}{2} \right)}^{x}}$
${{\log }_{\dfrac{1}{2}}}y=x$
$x={{\log }_{{{2}^{-1}}}}y$
$x=-{{\log }_{2}}y$
So the graph, $y={{\left( \dfrac{1}{2} \right)}^{x}}$ the inverse function of its is, $x=-{{\log }_{2}}y$
Inverse function means we have to reverse that original function or equation. For example, if $f$ is from $a$ to $b$ them its inverse will be ${{f}^{-1}}$ from $b$ to $a$.
i.e. $f(x)=b$ and ${{f}^{-1}}\left( b \right)=a$
First write the equation as it is given in the question. Then interchange the variable on both sides and take the logarithm on both sides.
Complete step-by-step answer:
The equation given in the question is as,
$y={{\left( \dfrac{1}{2} \right)}^{x}}...(i)$
Now, let’s interchange the variable so that we get,
$x={{\left( \dfrac{1}{2} \right)}^{y}}...(ii)$
Then we have to solve for $y.$
Now, take logarithm on both side of equation $(i)$ with base $\dfrac{1}{2}$
${{\log }_{\dfrac{1}{2}}}x={{\log }_{\dfrac{1}{2}}}{{\left( \dfrac{1}{2} \right)}^{y}}$
${{\log }_{\dfrac{1}{2}}}=y$
We can write ${{\log }_{\dfrac{1}{2}}}$ as ${{\log }_{2}}-1$
But according to logarithm rules.
${{\log }_{a}}-1=-{{\log }_{a}}$
$y=-{{\log }_{2}}x$
So, the graph of $y={{\left( \dfrac{1}{2} \right)}^{x}}$ and the inverse function of it is $y=-{{\log }_{2}}x$
Additional Information:
Let, $f$ is a function of $1-1$ along with domain $A$ and range $B.$
Then, the inverse function of above function will be as follows with domain \[B\] and range $A,$
The inverse of $f$ is denoted as ${{f}^{-1}}$
${{f}^{-1}}\left( y \right)=x$
Then $f(x)=y$ for any $y\in B$
Remember that inverse functions are only applicable for $1-1$ functions.
For example, if $h\left( -5 \right)=8$ then find ${{h}^{-1}}\left( 8 \right)=?$
Given, $h=\left( -5 \right)=8$
Let, $h\left( x \right)$ is the graph where $\left( -5,8 \right)$ points are located.
Now, inverse the graph $h(x)$ i.e. ${{h}^{-1}}\left( x \right)$ then the points on it will also get inverse i.e. $\left( 8,-5 \right)$on the graph.
So, when $h(x)=\left( -5,8 \right)$ after inversing ${{h}^{-1}}\left( x \right)=\left( 8,-5 \right)$
From above, we can say that,
If $h\left( -5 \right)=8$
Then, ${{h}^{-1}}\left( 8 \right)=-5$
Note:
Here we are taking the logarithm rule for simplifying the equation. We also interchange the variables which are not necessary. We can also solve it without interchanging the variables.
Then the solution will becomes,
After taking logarithm on both sides of the equation $(i)$.
${{\log }_{\dfrac{1}{2}}}y={{\log }_{\dfrac{1}{2}}}{{\left( \dfrac{1}{2} \right)}^{x}}$
${{\log }_{\dfrac{1}{2}}}y=x$
$x={{\log }_{{{2}^{-1}}}}y$
$x=-{{\log }_{2}}y$
So the graph, $y={{\left( \dfrac{1}{2} \right)}^{x}}$ the inverse function of its is, $x=-{{\log }_{2}}y$
Recently Updated Pages
Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Physics: Engaging Questions & Answers for Success

Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Chemistry: Engaging Questions & Answers for Success

Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Accountancy: Engaging Questions & Answers for Success

Trending doubts
Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

Proton was discovered by A Thomson B Rutherford C Chadwick class 11 chemistry CBSE

What is the purity of pure gold in carat class 11 chemistry CBSE

Can anyone list 10 advantages and disadvantages of friction

Lysosomes are known as suicidal bags of cell why class 11 biology CBSE

Derive the equation of motion by the calculus method class 11 physics CBSE

