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

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

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

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

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ln (x) - x = - 2

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

Decimal Notation

x = 3.14619 \dots, x = 0.15859 \dots

Solution steps

ln (x) - x = - 2

Move

x

to the right side

ln (x) = - 2 + x

f (x) = g (x),

then

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

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

Simplify

e^{l n (x)} : x

x = e^{- 2 + x}

Solve

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

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

Verify Solutions:

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

True

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

True

The solutions are

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

ln (x + 5) = 2 + ln (x)

lo g_{0.5}^{2} (1 - x^{2}) = 4

1.5 = - lo g_{10} (x)

3 lo g_{2} (x) = 12

ln (x + 2) = 11

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