Finding the inverse function of y = 1 + ln (x + 2)

Finding the inverse function of y = 1 + ln (x + 2)

y=1+ln(x+2)
y-1=ln(x+2)
x+2=e^(y-1)
x=e^(y-1)-2
So the inverse function of y = 1 + ln (x + 2) is y = e ^ (x-1) - 2

The inverse function of the function y = ln (x-1) is______ .

∵y=ln（x-1）
∴x=ey+1（y∈R），
The inverse function of the function y = ln (x-1) is y = ex + 1 (x ∈ R)
So the answer is: y = ex + 1 (x ∈ R)

Find the inverse function of y = ln (x + √ (x ^ 2 + 1))

y=ln(x+√(x^2+1))
x+(x^2+1)^(1/2)=e^y
(x^2+1)^(1/2)=e^y-x
x^2+1=e^2y-2xe^y+x^2
2xe^y=e^2y-1
x=(e^y)/2-[e^(-y)]/2=[e^y-e^(-y)]/2
Inverse function: y = [e ^ x-e ^ (- x)] / 2

The inverse function of the function y = ln (2x + 1) (x > 0) is

e^y=2x+1
x=(e^y-1)/2
X>0
2x+1>1
Y>0
So y = (e ^ x-1) / 2, x > 0

Why are y = 1 / 2 * [the x power of e] and y = ln (2x) inverse functions I don't know what happened

y=1/2*e^x
e^x=2y
x=ln(2y)
So the inverse function is y = ln (2x)

Find the inverse function of the function f (x) = 1-LN (2x + 1)

Let y = f (x) = 1-LN (2x + 1)
Then ln (2x + 1) = 1-y
So 2x + 1 = e ^ (1-y)
So x = [e ^ (1-y) - 1] / 2
If x and y are exchanged, then the inverse function f ^ - 1 (x) = [e ^ (1-x) - 1] / 2

The inverse function of the function f (x) = ln (x-1) (x > 1) is ()

f(x)=ln(x-1)
x-1=e^f(x)
x=e^f(x)+1
According to custom
Change y to X and X to y
f(x)=e^x+1

The inverse function of the function f (x) = 1 / ln (x-1) is:

y=1/ln(x-1), x>1, y0
1/y=ln(x-1)
x-1=e^(1/y)
x=1+e^(1/y)
The inverse function is: y = 1 + e ^ (1 / x), x0

Inverse function y = f (x) = ln (1-e ^ - x) Y = f (x) = ln (1-e ^ - x)

Y = f (x) = ln (1-e ^ - x)
x=ln(1-e^-y)
e^x=1-e^-y
e^-y=1-e^x
-y=ln(1-e^x)
y=-ln(1-e^x)

Find the inverse function of the function f (x) = ln (x-1) (x > 1)

Y = ln (x-1) e ^ y = X-1 x = e ^ y + 1 the inverse function is f (x) = e ^ y + 1