Solve for x natural log of x+ natural log of x+2=4

$\ln (x) + \ln (x + 2) = 4$

Step 1

Simplify the left side.

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

$\ln (x (x + 2)) = 4$

Apply the distributive property.

$\ln (x \cdot x + x \cdot 2) = 4$

Simplify the expression.

Multiply $x$ by $x$ .

$\ln (x^{2} + x \cdot 2) = 4$

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

$\ln (x^{2} + 2 x) = 4$

Step 2

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

$e^{\ln (x^{2} + 2 x)} = e^{4}$

Step 3

Rewrite $\ln (x^{2} + 2 x) = 4$ 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 $b^{y} = x$ .

$e^{4} = x^{2} + 2 x$

Step 4

Solve for $x$ .

Rewrite the equation as $x^{2} + 2 x = e^{4}$ .

$x^{2} + 2 x = e^{4}$

Subtract $e^{4}$ from both sides of the equation.

$x^{2} + 2 x - e^{4} = 0$

Use the quadratic formula to find the solutions.

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

Substitute the values $a = 1$ , $b = 2$ , and $c = - e^{4}$ into the quadratic formula and solve for $x$ .

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

Simplify.

Simplify the numerator.

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

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

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

Multiply $- 4$ by $1$ .

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

Multiply $- 4$ by $- 1$ .

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

Rewrite $4 + 4 e^{4}$ as $2^{2} (1 + e^{4})$ .

Factor $4$ out of $4$ .

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

Factor $4$ out of $4 e^{4}$ .

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

Factor $4$ out of $4 (1) + 4 (e^{4})$ .

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

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

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

Pull terms out from under the radical.

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

Multiply $2$ by $1$ .

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

Simplify $\frac{- 2 \pm 2 \sqrt{1 + e^{4}}}{2}$ .

$x = - 1 \pm \sqrt{1 + e^{4}}$

The final answer is the combination of both solutions.

$x = - 1 + \sqrt{1 + e^{4}}, - 1 - \sqrt{1 + e^{4}}$

Step 5

Exclude the solutions that do not make $\ln (x) + \ln (x + 2) = 4$ true.

$x = - 1 + \sqrt{1 + e^{4}}$

Step 6

The result can be shown in multiple forms.

Exact Form:

$x = - 1 + \sqrt{1 + e^{4}}$

Decimal Form:

$x = 6.45641670 \dots$

Solve for x natural log of x+ natural log of x+2=4

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