Calculus Examples

$\ln (x) + \ln (x + 2) = 0$

Step 1

Simplify the left side.

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Step 1.1

Use the product property of logarithms, $\log_{b} (x) + \log_{b} (y) = \log_{b} (x y)$ .

$\ln (x (x + 2)) = 0$

Step 1.2

Apply the distributive property.

$\ln (x \cdot x + x \cdot 2) = 0$

Step 1.3

Simplify the expression.

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Step 1.3.1

Multiply $x$ by $x$ .

$\ln (x^{2} + x \cdot 2) = 0$

Step 1.3.2

Move $2$ to the left of $x$ .

$\ln (x^{2} + 2 x) = 0$

Step 2

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

$e^{\ln (x^{2} + 2 x)} = e^{0}$

Step 3

Rewrite $\ln (x^{2} + 2 x) = 0$ 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 $b^{y} = x$ .

$e^{0} = x^{2} + 2 x$

Step 4

Solve for $x$ .

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Step 4.1

Rewrite the equation as $x^{2} + 2 x = e^{0}$ .

$x^{2} + 2 x = e^{0}$

Step 4.2

Anything raised to $0$ is $1$ .

$x^{2} + 2 x = 1$

Step 4.3

Subtract $1$ from both sides of the equation.

$x^{2} + 2 x - 1 = 0$

Step 4.4

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 4.5

Substitute the values $a = 1$ , $b = 2$ , and $c = - 1$ into the quadratic formula and solve for $x$ .

$\frac{- 2 \pm \sqrt{2^{2} - 4 \cdot (1 \cdot - 1)}}{2 \cdot 1}$

Step 4.6

Simplify.

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Step 4.6.1

Simplify the numerator.

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Step 4.6.1.1

Raise $2$ to the power of $2$ .

$x = \frac{- 2 \pm \sqrt{4 - 4 \cdot 1 \cdot - 1}}{2 \cdot 1}$

Step 4.6.1.2

Multiply $- 4 \cdot 1 \cdot - 1$ .

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Step 4.6.1.2.1

Multiply $- 4$ by $1$ .

$x = \frac{- 2 \pm \sqrt{4 - 4 \cdot - 1}}{2 \cdot 1}$

Step 4.6.1.2.2

Multiply $- 4$ by $- 1$ .

$x = \frac{- 2 \pm \sqrt{4 + 4}}{2 \cdot 1}$

Step 4.6.1.3

Add $4$ and $4$ .

$x = \frac{- 2 \pm \sqrt{8}}{2 \cdot 1}$

Step 4.6.1.4

Rewrite $8$ as $2^{2} \cdot 2$ .

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Step 4.6.1.4.1

Factor $4$ out of $8$ .

$x = \frac{- 2 \pm \sqrt{4 (2)}}{2 \cdot 1}$

Step 4.6.1.4.2

Rewrite $4$ as $2^{2}$ .

$x = \frac{- 2 \pm \sqrt{2^{2} \cdot 2}}{2 \cdot 1}$

Step 4.6.1.5

Pull terms out from under the radical.

$x = \frac{- 2 \pm 2 \sqrt{2}}{2 \cdot 1}$

Step 4.6.2

Multiply $2$ by $1$ .

$x = \frac{- 2 \pm 2 \sqrt{2}}{2}$

Step 4.6.3

Simplify $\frac{- 2 \pm 2 \sqrt{2}}{2}$ .

$x = - 1 \pm \sqrt{2}$

Step 4.7

The final answer is the combination of both solutions.

$x = - 1 + \sqrt{2}, - 1 - \sqrt{2}$

Step 5

Exclude the solutions that do not make $\ln (x) + \ln (x + 2) = 0$ true.

$x = - 1 + \sqrt{2}$

Step 6

The result can be shown in multiple forms.

Exact Form:

$x = - 1 + \sqrt{2}$

Decimal Form:

$x = 0.41421356 \dots$