What is the answer to ln(x)=2+ln(1-x) ?

The answer to ln(x)=2+ln(1-x) is x=(e^2)/(1+e^2)

Solve for x: ln(x)=2+ln(1-x)

The solution is x=(e^2)/(1+e^2)

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

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

Decimal Notation

x = 0.88079 \dots

Solution steps

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

Apply log rules

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

Solve

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

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

Verify Solutions:

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

True

The solution is

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

5 + lo g_{2} (- 5 x) = 9

lo g_{10} (x) = - 4

ln (x + 2) = 1

ln (x^{2}) = 16

lo g_{10} (x) = 0.3

What is the answer to ln(x)=2+ln(1-x) ?
The answer to ln(x)=2+ln(1-x) is x=(e^2)/(1+e^2)
Solve for x: ln(x)=2+ln(1-x)
The solution is x=(e^2)/(1+e^2)