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

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_{- 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}})

lo g_{3} (x + 1) = 2

\frac{lo g _{5} ( x ) - 4}{lo g _{5} ( x ) + 3} = 0

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

lo g_{4} (2^{x} + 48) = x - 1

lo g_{4} (x^{2} + 3 x - 1) = lo g_{4} (5 x - 1)

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))