Calculus Examples

1 + ln (x) = 0

$1 + \ln (x) = 0$

Step 1

Subtract

1

$1$ from both sides of the equation.

ln (x) = - 1

$\ln (x) = - 1$

Step 2

To solve for

x

$x$ , rewrite the equation using properties of logarithms.

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

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

Step 3

Rewrite

ln (x) = - 1

$\ln (x) = - 1$ in exponential form using the definition of a logarithm. If

x

$x$ and

b

$b$ are positive real numbers and

b \neq 1

$b \neq 1$ , then

{log}_{b} (x) = y

$\log_{b} (x) = y$ is equivalent to

b^{y} = x

$b^{y} = x$ .

e^{- 1} = x

$e^{- 1} = x$

Step 4

Solve for

x

$x$ .

Tap for more steps...

Step 4.1

Rewrite the equation as

x = e^{- 1}

$x = e^{- 1}$ .

x = e^{- 1}

$x = e^{- 1}$

Step 4.2

Rewrite the expression using the negative exponent rule

b^{- n} = \frac{1}{b^{n}}

$b^{- n} = \frac{1}{b^{n}}$ .

x = \frac{1}{e}

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

x = \frac{1}{e}

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

Step 5