What is the answer to solvefor u,ln(u)-ln(1-u)=t+C ?

The answer to solvefor u,ln(u)-ln(1-u)=t+C is u=(e^{t+C})/(1+e^{t+C)}

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solvefor u,ln(u)-ln(1-u)=t+C

solve for

u, ln (u) - ln (1 - u) = t + C

u = \frac{e ^{t + C}}{1 + e ^{t + C}}

Solution steps

ln (u) - ln (1 - u) = t + C

Add

ln (1 - u)

to both sides

ln (u) - ln (1 - u) + ln (1 - u) = t + C + ln (1 - u)

Simplify

ln (u) = t + C + ln (1 - u)

Apply log rules

u = e^{t + C} (1 - u)

Solve

u = e^{t + C} (1 - u) : u = \frac{e ^{t + C}}{1 + e ^{t + C}}

u = \frac{e ^{t + C}}{1 + e ^{t + C}}

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

3 ln (6 x - 5) + 4 = - 9

x, 2 ln (7 x) = - 2

60 = 20 lo g_{10} (\frac{x}{100})

125 = 10 lo g_{10} (x)

What is the answer to solvefor u,ln(u)-ln(1-u)=t+C ?
The answer to solvefor u,ln(u)-ln(1-u)=t+C is u=(e^{t+C})/(1+e^{t+C)}