What is the answer to solvefor y,ln(y)-ln(1-y)=x ?

The answer to solvefor y,ln(y)-ln(1-y)=x is y=(e^x)/(1+e^x)

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solvefor y,ln(y)-ln(1-y)=x

solve for

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

y = \frac{e ^{x}}{1 + e ^{x}}

Solution steps

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

Add

ln (1 - y)

to both sides

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

Simplify

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

Apply log rules

y = e^{x} (1 - y)

Solve

y = e^{x} (1 - y) : y = \frac{e ^{x}}{1 + e ^{x}}

y = \frac{e ^{x}}{1 + e ^{x}}

Verify Solutions:

y = \frac{e ^{x}}{1 + e ^{x}}

True

The solution is

y = \frac{e ^{x}}{1 + e ^{x}}

lo g_{2} (2 x - 4) = 4

lo g_{2} (x + 1) + lo g_{2} (x - 1) = 4

ln (\frac{( t - 63 )}{125}) = - 0.07 t

2 ln (4 x + 1) + 14 = 20

7.5 = - lo g_{10} (h)

What is the answer to solvefor y,ln(y)-ln(1-y)=x ?
The answer to solvefor y,ln(y)-ln(1-y)=x is y=(e^x)/(1+e^x)