What is the answer to 0.41667=-0.25log_{e}(1-c) ?

The answer to 0.41667=-0.25log_{e}(1-c) is c=(-1+e^{\frac{0.41667)/(0.25)}}{e^{(0.41667)/(0.25)}}

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0.41667=-0.25log_{e}(1-c)

0.41667 = - 0.25 lo g_{e} (1 - c)

c = \frac{- 1 + e ^{\frac{0.41667}{0.25}}}{e ^{\frac{0.41667}{0.25}}}

Decimal

c = 0.81112 \dots

Solution steps

0.41667 = - 0.25 lo g_{e} (1 - c)

Switch sides

- 0.25 lo g_{e} (1 - c) = 0.41667

Divide both sides by

- 0.25

lo g_{e} (1 - c) = - \frac{0.41667}{0.25}

Apply log rules

1 - c = \frac{1}{e ^{\frac{0.41667}{0.25}}}

Solve

1 - c = \frac{1}{e ^{\frac{0.41667}{0.25}}} : c = \frac{- 1 + e ^{\frac{0.41667}{0.25}}}{e ^{\frac{0.41667}{0.25}}}

c = \frac{- 1 + e ^{\frac{0.41667}{0.25}}}{e ^{\frac{0.41667}{0.25}}}

Verify Solutions:

c = \frac{- 1 + e ^{\frac{0.41667}{0.25}}}{e ^{\frac{0.41667}{0.25}}}

True

The solution is

c = \frac{- 1 + e ^{\frac{0.41667}{0.25}}}{e ^{\frac{0.41667}{0.25}}}

lo g_{10} (x) = x - 3

lo g_{x} (2) = - 2

lo g_{x} (2) = - 1

lo g_{10} (x - 15) + lo g_{10} (x) = 2

c + 5 ln (- 1 + c) = 0.540302

What is the answer to 0.41667=-0.25log_{e}(1-c) ?
The answer to 0.41667=-0.25log_{e}(1-c) is c=(-1+e^{\frac{0.41667)/(0.25)}}{e^{(0.41667)/(0.25)}}