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

How do you find the inverse of y=ln⁡(xx−1)y=\ln \left( \dfrac{x}{x-1} \right) ?

Explanation

Solution

We are given a function as y=ln⁡(xx−1)y=\ln \left( \dfrac{x}{x-1} \right), we are asked to find the inverse of it, we will start by learn what are called the inverse of a function then we will following step by step certain to find its inverse, we will need the knowledge that Log and exponent are opposite to each other they are inverse of one another, we will simplify and solve for one answer.

Complete step by step solution:
We are given a function as y=ln⁡(xx−1)y=\ln \left( \dfrac{x}{x-1} \right)we have to find the value of its inverse.
Now, before we move forward, remember for a function say f(x)f\left( x \right) , its inverse is some function represented as f−1(x){{f}^{-1}}\left( x \right) such that their composition is always identify function that is –
(fof)−1(x)=(f−1of)(x)=I(x)=x{{\left( fof \right)}^{-1}}\left( x \right)=\left( {{f}^{-1}}of \right)\left( x \right)=I\left( x \right)=x
Now to find the inverse of any function f(x)f\left( x \right) we follow these steps –
Step 1: Mark the function f(x)f\left( x \right) as ‘y’
Step 2: Replace all ‘x’ as ‘y’ and all ‘y’ as ‘x’
Step 3: We will now solve for the value of ‘y’.
Step 4: Once we get value of ‘y’, replace ‘y’ f−1(x){{f}^{-1}}\left( x \right)
We will follow these steps to get to our required solution.
Now we have y=ln⁡(xx−1)y=\ln \left( \dfrac{x}{x-1} \right)
As we can see that it is already in term of ‘y’ so step 1 is already completed.
Now, by step 2, we change ‘x’ as ‘y’ and ‘y’ as ‘x’.
So, we get –
y=ln⁡(yy−1)y=\ln \left( \dfrac{y}{y-1} \right)………………………………….. (1)
Now we will follow step (3) and solve for the value of ‘y’.
To do so we will use that log and exponential one inverse so e(ln⁡(x))=ln⁡(e(x))=xe\left( \ln \left( x \right) \right)=\ln \left( e\left( x \right) \right)=x
So, we apply exponential on both side of equation (1), we get –
ex=eln⁡(yy−1){{e}^{x}}={{e}^{\ln \left( \dfrac{y}{y-1} \right)}}
As eln⁡(x)=x{{e}^{\ln(x) }}=x,
So we get –
ex=yy−1{{e}^{x}}=\dfrac{y}{y-1}
Now, we simplify we get –
(y−1)ex=y\left( y-1 \right){{e}^{x}}=y opening brackets, we get –
yex−ex=yy{{e}^{x}}-{{e}^{x}}=y
Subtracting yexy{{e}^{x}} on both side, we get –
yex−ex−yex=y−yexy{{e}^{x}}-{{e}^{x}}-y{{e}^{x}}=y-y{{e}^{x}}
So, −ex=y−yex-{{e}^{x}}=y-y{{e}^{x}}
Now taking ‘y’ as common, we get –
−ex=y(1−ex)-{{e}^{x}}=y\left( 1-{{e}^{x}} \right)
Now dividing both sides by 1−ex1-{{e}^{x}} ,
We get –
y=−ex1−exy=\dfrac{-{{e}^{x}}}{1-{{e}^{x}}}
By simplifying, we get –
y=ex−(1−ex)=exex−1y=\dfrac{{{e}^{x}}}{-\left( 1-{{e}^{x}} \right)}=\dfrac{{{e}^{x}}}{{{e}^{x}}-1}
Now we follow step 4, we replace ‘y’ as inverse if function so,
f−1(x)=exex−1{{f}^{-1}}\left( x \right)=\dfrac{{{e}^{x}}}{{{e}^{x}}-1}

Hence, inverse of y=ln⁡(xx−1)y=\ln \left( \dfrac{x}{x-1} \right) is exex−1\dfrac{{{e}^{x}}}{{{e}^{x}}-1}

Note: We can cross check our solution by checking f(f−1(x))=f−1(x)=exex−1f\left( {{f}^{-1}}\left( x \right) \right)={{f}^{-1}}\left( x \right)=\dfrac{{{e}^{x}}}{{{e}^{x}}-1}
Now we first find (fof−1)(x)\left( fo{{f}^{-1}} \right)\left( x \right)
So, f(f−1(x))=ln⁡(exex−1exex−1−1)f\left( {{f}^{-1}}\left( x \right) \right)=\ln \left( \dfrac{\dfrac{{{e}^{x}}}{{{e}^{x}}-1}}{\dfrac{{{e}^{x}}}{{{e}^{x}}-1}-1} \right)
By simplifying, we get –
ln⁡(exex−1ex−ex+1ex−1)\ln \left( \dfrac{\dfrac{{{e}^{x}}}{{{e}^{x}}-1}}{\dfrac{{{e}^{x}}-{{e}^{x}}+1}{{{e}^{x}}-1}} \right)
Cancelling like term, we get –
=ln⁡(ex)=\ln \left( {{e}^{x}} \right)
As we know ln⁡(ex)=x\ln \left( {{e}^{x}} \right)=x
So we get –
=x=x
Hence, (fof−1)(x)=x\left( fo{{f}^{-1}} \right)\left( x \right)=x
Now we check (f−1of)(x)\left( {{f}^{-1}}of \right)\left( x \right)
So, (exex−1)o(ln⁡(xx−1))=eln⁡(xx−1)eln⁡(xx−1)−1\left( \dfrac{{{e}^{x}}}{{{e}^{x}}-1} \right)o\left( \ln \left( \dfrac{x}{x-1} \right) \right)=\dfrac{{{e}^{\ln \left( \dfrac{x}{x-1} \right)}}}{{{e}^{\ln \left( \dfrac{x}{x-1} \right)}}-1}
As eln⁡(x)=x{{e}^{\ln \left( x \right)=x}} so, eln⁡(nn−1)=nn−1{{e}^{\ln \left( \dfrac{n}{n-1} \right)=\dfrac{n}{n-1}}}
Hence, we get –
=xx−1xx−1−1=\dfrac{\dfrac{x}{x-1}}{\dfrac{x}{x-1}-1}
By simplifying we get –
=xx−1x−x+1x−1=x=\dfrac{\dfrac{x}{x-1}}{\dfrac{x-x+1}{x-1}}=x
So, we get –
(fof−1)(x)=x\left( fo{{f}^{-1}} \right)\left( x \right)=x
Hence we can write inverse of ln⁡(xx−1)\ln \left( \dfrac{x}{x-1} \right) is exex−1\dfrac{{{e}^{x}}}{{{e}^{x}}-1} .