What is the answer to ln|x|=x-2 ?

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

Solve for x: ln|x|=x-2

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

ln ∣ x ∣ = x - 2

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

Decimal Notation

x = - 0.12002 \dots, x = 0.15859 \dots, x = 3.14619 \dots

Solution steps

ln ∣ x ∣ = x - 2

f (x) = g (x),

then

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

e^{l n ∣ x ∣} = e^{x - 2}

Simplify

e^{l n ∣ x ∣} : ∣ x ∣

∣ x ∣ = e^{x - 2}

Solve

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

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

Verify Solutions:

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

True

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

True

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

True

The solutions are

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

lo g_{36} (x) = \frac{1}{2}

lo g_{125} (x) = \frac{1}{3}

lo g_{x} (8) = - 3

lo g_{10} (3 x) = 3

lo g_{2} (x) + lo g_{2} (2 x - 7) = 2

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