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

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

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

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

x = \frac{e - 1}{1 + e}

Decimal Notation

x = 0.46211 \dots

Solution steps

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

Add

ln (1 - x)

to both sides

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

Simplify

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

Apply log rules

1 + x = e (1 - x)

Solve

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

x = \frac{e - 1}{1 + e}

Verify Solutions:

x = \frac{e - 1}{1 + e}

True

The solution is

x = \frac{e - 1}{1 + e}

lo g_{10} (x) = - 1

lo g_{3} (x) + lo g_{3} (x + 1) = 8

0.1 = lo g_{10} (x)

T, - lo g_{10} (T) = 6.016

lo g_{2} (2 x) = 2

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