What is the answer to solvefor y,x=ln(2-y)-ln(1-y) ?

The answer to solvefor y,x=ln(2-y)-ln(1-y) is y=(2-e^x)/(-e^x+1) {x>0}

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

solve for

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

y = \frac{2 - e ^{x}}{- e ^{x} + 1} {x > 0}

Solution steps

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

Add

ln (1 - y)

to both sides

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

Simplify

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

Apply log rules

e^{x} (1 - y) = 2 - y

Solve

e^{x} (1 - y) = 2 - y : y = \frac{2 - e ^{x}}{- e ^{x} + 1}

y = \frac{2 - e ^{x}}{- e ^{x} + 1}

Verify Solutions:

y = \frac{2 - e ^{x}}{- e ^{x} + 1} {x > 0}

The solution is

y = \frac{2 - e ^{x}}{- e ^{x} + 1} {x > 0}

lo g_{e} (y + 3) = 19

x^{2} = x \cdot lo g_{10} (x)

\frac{1}{y} = ln (y)

lo g_{2} (\frac{3}{4} x + 4) = 6

x, ln (2 x + 3) = 7

What is the answer to solvefor y,x=ln(2-y)-ln(1-y) ?
The answer to solvefor y,x=ln(2-y)-ln(1-y) is y=(2-e^x)/(-e^x+1) {x>0}