Find Inverse of f(x) = ln(x-1)

In summary, we discussed the inverse function of f(x) = ln(x-1), which is f^-1(x) = e^x + 1. To find the inverse, we switch the variables and solve for y, resulting in y = e^x + 1. The domain of f(x) is all real numbers greater than 1, and the range is all real numbers. The inverse is a one-to-one function and can also be written as f^-1(x) = log(e, x+1).
  • #1
spartas
7
0
find the inverse function of
f(x)=ln(x-1), x>1
 
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  • #2
spartas said:
find the inverse function of
f(x)=ln(x-1), x>1

You set $y=f(x)$. Then $y= \ln (x-1) \Rightarrow e^y=x-1 \Rightarrow x=e^y+1$.

What can we deduce from that?
 

FAQ: Find Inverse of f(x) = ln(x-1)

1. What is the inverse function of f(x) = ln(x-1)?

The inverse function of f(x) = ln(x-1) is f^-1(x) = e^x + 1.

2. How do you find the inverse of f(x) = ln(x-1)?

To find the inverse of f(x) = ln(x-1), switch the x and y variables and solve for y. In this case, it would be x = ln(y-1). Then, rewrite the equation in exponential form to get y = e^x + 1, which is the inverse function.

3. What is the domain and range of f(x) = ln(x-1)?

The domain of f(x) = ln(x-1) is all real numbers greater than 1, since the natural logarithm of 0 or a negative number is undefined. The range is all real numbers.

4. Is the inverse of f(x) = ln(x-1) a one-to-one function?

Yes, the inverse of f(x) = ln(x-1) is a one-to-one function because each input has a unique output and vice versa. This can be seen by graphing both functions and observing that they are reflections of each other across the line y = x.

5. Can the inverse of f(x) = ln(x-1) be written in a different form?

Yes, the inverse of f(x) = ln(x-1) can also be written as f^-1(x) = log(e, x+1), where log(e, x) represents the natural logarithm with base e. These two forms are equivalent and can be used interchangeably.

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