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

The answer to x=-ln(2-x) is x=\W_{-1}(-1/(e^2))+2,x=\W_{0}(-1/(e^2))+2

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

The solution is x=\W_{-1}(-1/(e^2))+2,x=\W_{0}(-1/(e^2))+2

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x = - ln (2 - x)

x = W_{- 1} (- \frac{1}{e ^{2}}) + 2, x = W_{0} (- \frac{1}{e ^{2}}) + 2

Decimal Notation

x = - 1.14619 \dots, x = 1.84140 \dots

Solution steps

x = - ln (2 - x)

Move

ln (2 - x)

to the left side

x + ln (2 - x) = 0

Move

x

to the right side

ln (2 - x) = - x

f (x) = g (x),

then

a^{f (x)} = a^{g (x)}

e^{l n (2 - x)} = e^{- x}

Simplify

e^{l n (2 - x)} : 2 - x

2 - x = e^{- x}

Solve

2 - x = e^{- x} : x = W_{- 1} (- \frac{1}{e ^{2}}) + 2, x = W_{0} (- \frac{1}{e ^{2}}) + 2

x = W_{- 1} (- \frac{1}{e ^{2}}) + 2, x = W_{0} (- \frac{1}{e ^{2}}) + 2

Verify Solutions:

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

True

, x = W_{0} (- \frac{1}{e ^{2}}) + 2

True

The solutions are

x = W_{- 1} (- \frac{1}{e ^{2}}) + 2, x = W_{0} (- \frac{1}{e ^{2}}) + 2

a, 1 = b lo g_{5} (6 + a)

ln (3 x + 4) = 6

ln (2 x + 1) + ln (x + 2) = 0

lo g_{2} (\frac{1}{x ^{2}}) = 3

5.5 = 6 + lo g_{10} (x)

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