What is the answer to log_{e}(x/(1-x))=e^2 ?

The answer to log_{e}(x/(1-x))=e^2 is x=(e^{e^2})/(1+e^{e^2)}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

lo g_{e} (\frac{x}{1 - x}) = e^{2}

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

Decimal

x = 0.99938 \dots

Solution steps

lo g_{e} (\frac{x}{1 - x}) = e^{2}

Apply log rules

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

Multiply both sides by

1 - x

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

Simplify

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

Solve

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

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

Verify Solutions:

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

True

The solution is

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

- 2.26 = lo g_{10} (x)

lo g_{16} (b) + lo g_{16} (b - 15) = 1

t, x = ln (4 t)

lo g_{2} (5 - 3 x) = 3

lo g_{3} (9^{x} - 2) = x

What is the answer to log_{e}(x/(1-x))=e^2 ?
The answer to log_{e}(x/(1-x))=e^2 is x=(e^{e^2})/(1+e^{e^2)}