What is the answer to -ln(1-p)=0.002 ?

The answer to -ln(1-p)=0.002 is p=(-1+e^{0.002})/(e^{0.002)}

Solve for p: -ln(1-p)=0.002

The solution is p=(-1+e^{0.002})/(e^{0.002)}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

- ln (1 - p) = 0.002

p = \frac{- 1 + e ^{0.002}}{e ^{0.002}}

Solution steps

- ln (1 - p) = 0.002

Divide both sides by

- 1

ln (1 - p) = - 0.002

Apply log rules

1 - p = \frac{1}{e ^{0.002}}

Solve

1 - p = \frac{1}{e ^{0.002}} : p = \frac{- 1 + e ^{0.002}}{e ^{0.002}}

p = \frac{- 1 + e ^{0.002}}{e ^{0.002}}

Verify Solutions:

p = \frac{- 1 + e ^{0.002}}{e ^{0.002}}

True

The solution is

p = \frac{- 1 + e ^{0.002}}{e ^{0.002}}

lo g_{6} (x - 2) = 4 + lo g_{6} (x + 2)

lo g_{4} (x) - lo g_{4} (x + 9) = - 1

ln (n^{2} + 12) = ln (- 9 n - 2)

lo g_{2} (x + 1) - 2 lo g_{2} (3) = 3

7.5 = 10.5 + lo g_{10} (x)

What is the answer to -ln(1-p)=0.002 ?
The answer to -ln(1-p)=0.002 is p=(-1+e^{0.002})/(e^{0.002)}
Solve for p: -ln(1-p)=0.002
The solution is p=(-1+e^{0.002})/(e^{0.002)}