Solve for x 0=1+ natural log of x

$0 = 1 + \ln (x)$

Rewrite the equation as $1 + \ln (x) = 0$ .

$1 + \ln (x) = 0$

Subtract $1$ from both sides of the equation.

$\ln (x) = - 1$

To solve for $x$ , rewrite the equation using properties of logarithms.

$e^{\ln (x)} = e^{- 1}$

Rewrite $\ln (x) = - 1$ in exponential form using the definition of a logarithm. If $x$ and $b$ are positive real numbers and $b \neq 1$ , then $\log_{b} (x) = y$ is equivalent to $b^{y} = x$ .

$e^{- 1} = x$

Solve for $x$ .

Rewrite the equation as $x = e^{- 1}$ .

$x = e^{- 1}$

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$x = \frac{1}{e}$

The result can be shown in multiple forms.

Exact Form:

$x = \frac{1}{e}$

Decimal Form:

$x = 0.36787944 \dots$

Solve for x 0=1+ natural log of x

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