What is the answer to ln(2-x)+2=0 ?

The answer to ln(2-x)+2=0 is x=(-1+2e^2)/(e^2)

Solve for x: ln(2-x)+2=0

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

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ln (2 - x) + 2 = 0

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

Decimal Notation

x = 1.86466 \dots

Solution steps

ln (2 - x) + 2 = 0

Move

2

to the right side

ln (2 - x) = - 2

Apply log rules

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

Solve

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

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

Verify Solutions:

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

True

The solution is

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

lo g_{x} (3) = \frac{1}{3}

- 1 = ln (x)

lo g_{e} (x) = ln (x)

4 ln (4 x) = ln (2 x)

\frac{1}{x} = ln (x)

What is the answer to ln(2-x)+2=0 ?
The answer to ln(2-x)+2=0 is x=(-1+2e^2)/(e^2)
Solve for x: ln(2-x)+2=0
The solution is x=(-1+2e^2)/(e^2)