in f(x)=ln(x/x-1) how would you write an exporession for f^-1(x), where f^-1(x) denotes the inverse of f(x)
I assume the correct expression is
y=f(x)=ln(x/(x-1))
taking inverse of ln(),
ey = x/(x-1)
(x-1)ey = x
x = ey / (e(y -1)
Therefore
f-1(x) = ex / (e(x -1)
erratum:
f-1(x) = ex/(ex-1)
To find the inverse function f^-1(x) of f(x), we need to interchange the roles of x and f(x) and solve for x.
Step 1: Replace f(x) with y. The equation becomes:
y = ln(x/(x-1))
Step 2: Interchange x and y:
x = ln(y/(y-1))
Step 3: Solve for y:
To isolate y, we need to undo the natural logarithm. Rewrite the equation in exponential form:
e^x = y/(y-1)
Step 4: Manipulate the equation to solve for y:
Multiply both sides of the equation by (y - 1) to eliminate the denominator:
(e^x)(y - 1) = y
Expand the equation:
e^xy - e^x = y
Rearrange the terms:
e^xy - y = e^x
Now, let's solve this equation for y to express the inverse function f^-1(x):
e^xy - y = e^x
e^xy - y + e^x = 0
This equation cannot be solved explicitly for y, but we can use numerical methods or approximation techniques to find the inverse function f^-1(x).