Solve for x natural log of x+ natural log of x^2=6

$\ln (x) + \ln (x^{2}) = 6$

Simplify the left side.

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

$\ln (x \cdot x^{2}) = 6$

Multiply $x$ by $x^{2}$ by adding the exponents.

Multiply $x$ by $x^{2}$ .

Raise $x$ to the power of $1$ .

$\ln (x \cdot x^{2}) = 6$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$\ln (x^{1 + 2}) = 6$

Add $1$ and $2$ .

$\ln (x^{3}) = 6$

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

$e^{\ln (x^{3})} = e^{6}$

Rewrite $\ln (x^{3}) = 6$ 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 $b^{y} = x$ .

$e^{6} = x^{3}$

Solve for $x$ .

Rewrite the equation as $x^{3} = e^{6}$ .

$x^{3} = e^{6}$

Subtract $e^{6}$ from both sides of the equation.

$x^{3} - e^{6} = 0$

Factor the left side of the equation.

Rewrite $e^{6}$ as ${(e^{2})}^{3}$ .

$x^{3} - {(e^{2})}^{3} = 0$

Since both terms are perfect cubes, factor using the difference of cubes formula, $a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2})$ where $a = x$ and $b = e^{2}$ .

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

Simplify.

Multiply the exponents in ${(e^{2})}^{2}$ .

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

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

Multiply $2$ by $2$ .

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

Reorder terms.

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

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x - e^{2} = 0$

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

Set $x - e^{2}$ equal to $0$ and solve for $x$ .

Set $x - e^{2}$ equal to $0$ .

$x - e^{2} = 0$

Add $e^{2}$ to both sides of the equation.

$x = e^{2}$

Set $e^{4} + x^{2} + e^{2} x$ equal to $0$ and solve for $x$ .

Set $e^{4} + x^{2} + e^{2} x$ equal to $0$ .

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

Solve $e^{4} + x^{2} + e^{2} x = 0$ for $x$ .

Use the quadratic formula to find the solutions.

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

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

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

Simplify.

Simplify the numerator.

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

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

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = e^{2}$ and $b = 2 \cdot 1 \cdot e^{2}$ .

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

Simplify.

Multiply $2$ by $1$ .

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

Add $e^{2}$ and $2 e^{2}$ .

$x = \frac{- e^{2} \pm \sqrt{3 e^{2} (e^{2} - (2 \cdot 1 \cdot e^{2}))}}{2 \cdot 1}$

Combine exponents.

Multiply $2$ by $1$ .

$x = \frac{- e^{2} \pm \sqrt{3 e^{2} (e^{2} - (2 \cdot e^{2}))}}{2 \cdot 1}$

Multiply $2$ by $- 1$ .

$x = \frac{- e^{2} \pm \sqrt{3 e^{2} (e^{2} - 2 e^{2})}}{2 \cdot 1}$

Subtract $2 e^{2}$ from $e^{2}$ .

$x = \frac{- e^{2} \pm \sqrt{3 e^{2} \cdot (- 1 e^{2})}}{2 \cdot 1}$

Combine exponents.

Factor out negative.

$x = \frac{- e^{2} \pm \sqrt{- (3 e^{2} e^{2})}}{2 \cdot 1}$

Multiply $e^{2}$ by $e^{2}$ by adding the exponents.

Move $e^{2}$ .

$x = \frac{- e^{2} \pm \sqrt{- (3 (e^{2} e^{2}))}}{2 \cdot 1}$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$x = \frac{- e^{2} \pm \sqrt{- (3 e^{2 + 2})}}{2 \cdot 1}$

Add $2$ and $2$ .

$x = \frac{- e^{2} \pm \sqrt{- (3 e^{4})}}{2 \cdot 1}$

Multiply $3$ by $- 1$ .

$x = \frac{- e^{2} \pm \sqrt{- 3 e^{4}}}{2 \cdot 1}$

Rewrite $- 3 e^{4}$ as ${(i e^{2})}^{2} \cdot 3$ .

Rewrite $- 3$ as $- 1 (3)$ .

$x = \frac{- e^{2} \pm \sqrt{- 1 \cdot 3 e^{4}}}{2 \cdot 1}$

Rewrite $- 1$ as $i^{2}$ .

$x = \frac{- e^{2} \pm \sqrt{i^{2} \cdot (3 e^{4})}}{2 \cdot 1}$

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

$x = \frac{- e^{2} \pm \sqrt{i^{2} \cdot (3 {(e^{2})}^{2})}}{2 \cdot 1}$

Move $3$ .

$x = \frac{- e^{2} \pm \sqrt{i^{2} {(e^{2})}^{2} \cdot 3}}{2 \cdot 1}$

Rewrite $i^{2} {(e^{2})}^{2}$ as ${(i e^{2})}^{2}$ .

$x = \frac{- e^{2} \pm \sqrt{{(i e^{2})}^{2} \cdot 3}}{2 \cdot 1}$

Pull terms out from under the radical.

$x = \frac{- e^{2} \pm i e^{2} \sqrt{3}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{- e^{2} \pm i e^{2} \sqrt{3}}{2}$

The final answer is the combination of both solutions.

$x = - \frac{e^{2} - i e^{2} \sqrt{3}}{2}, - \frac{e^{2} + i e^{2} \sqrt{3}}{2}$

The final solution is all the values that make $(x - e^{2}) (e^{4} + x^{2} + e^{2} x) = 0$ true.

$x = e^{2}, - \frac{e^{2} - i e^{2} \sqrt{3}}{2}, - \frac{e^{2} + i e^{2} \sqrt{3}}{2}$

Solve for x natural log of x+ natural log of x^2=6

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