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Question: How do you find the inverse of \(y={{\left( \dfrac{1}{2} \right)}^{x}}?\)...

How do you find the inverse of y=(12)x?y={{\left( \dfrac{1}{2} \right)}^{x}}?

Explanation

Solution

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 ff is from aa to bb them its inverse will be f−1{{f}^{-1}} from bb to aa.
i.e. f(x)=bf(x)=b and f−1(b)=a{{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=(12)x...(i)y={{\left( \dfrac{1}{2} \right)}^{x}}...(i)
Now, let’s interchange the variable so that we get,
x=(12)y...(ii)x={{\left( \dfrac{1}{2} \right)}^{y}}...(ii)
Then we have to solve for y.y.
Now, take logarithm on both side of equation (i)(i) with base 12\dfrac{1}{2}
log⁡12x=log⁡12(12)y{{\log }_{\dfrac{1}{2}}}x={{\log }_{\dfrac{1}{2}}}{{\left( \dfrac{1}{2} \right)}^{y}}
log⁡12=y{{\log }_{\dfrac{1}{2}}}=y
We can write log⁡12{{\log }_{\dfrac{1}{2}}} as log⁡2−1{{\log }_{2}}-1
But according to logarithm rules.
log⁡a−1=−log⁡a{{\log }_{a}}-1=-{{\log }_{a}}
y=−log⁡2xy=-{{\log }_{2}}x

So, the graph of y=(12)xy={{\left( \dfrac{1}{2} \right)}^{x}} and the inverse function of it is y=−log⁡2xy=-{{\log }_{2}}x

Additional Information:
Let, ff is a function of 1−11-1 along with domain AA and range B.B.
Then, the inverse function of above function will be as follows with domain BB and range A,A,
The inverse of ff is denoted as f−1{{f}^{-1}}
f−1(y)=x{{f}^{-1}}\left( y \right)=x
Then f(x)=yf(x)=y for any y∈By\in B
Remember that inverse functions are only applicable for 1−11-1 functions.
For example, if h(−5)=8h\left( -5 \right)=8 then find h−1(8)=?{{h}^{-1}}\left( 8 \right)=?
Given, h=(−5)=8h=\left( -5 \right)=8
Let, h(x)h\left( x \right) is the graph where (−5,8)\left( -5,8 \right) points are located.
Now, inverse the graph h(x)h(x) i.e. h−1(x){{h}^{-1}}\left( x \right) then the points on it will also get inverse i.e. (8,−5)\left( 8,-5 \right)on the graph.
So, when h(x)=(−5,8)h(x)=\left( -5,8 \right) after inversing h−1(x)=(8,−5){{h}^{-1}}\left( x \right)=\left( 8,-5 \right)
From above, we can say that,
If h(−5)=8h\left( -5 \right)=8
Then, h−1(8)=−5{{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)(i).
log⁡12y=log⁡12(12)x{{\log }_{\dfrac{1}{2}}}y={{\log }_{\dfrac{1}{2}}}{{\left( \dfrac{1}{2} \right)}^{x}}
log⁡12y=x{{\log }_{\dfrac{1}{2}}}y=x
x=log⁡2−1yx={{\log }_{{{2}^{-1}}}}y
x=−log⁡2yx=-{{\log }_{2}}y
So the graph, y=(12)xy={{\left( \dfrac{1}{2} \right)}^{x}} the inverse function of its is, x=−log⁡2yx=-{{\log }_{2}}y