What is the inverse of y=ln(x-1)-ln(2x+1) ?

The inverse of y=ln(x-1)-ln(2x+1) is (-1-e^x)/(2e^x-1)

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inverse of y=ln(x-1)-ln(2x+1)

inverse

y = ln (x - 1) - ln (2 x + 1)

\frac{- 1 - e ^{x}}{2 e ^{x} - 1}

Solution steps

y = ln (x - 1) - ln (2 x + 1)

Replace

x

with

y

x = ln (y - 1) - ln (2 y + 1)

Solve for

y, x = ln (y - 1) - ln (2 y + 1)

\frac{- 1 - e ^{x}}{2 e ^{x} - 1}

\frac{1}{( x ^{2} - 2 x + 2 ) ^{2}}

y = \frac{1}{40} e^{0.1 x} + 30

f (x) = (x - 19)^{2}, [19, \infty]

f (x) = \frac{2}{- x + 4}

9 sin (x)

What is the inverse of y=ln(x-1)-ln(2x+1) ?
The inverse of y=ln(x-1)-ln(2x+1) is (-1-e^x)/(2e^x-1)