Solve for x natural log of e^( natural log of x)+ natural log of e^( natural log of x^2)=2 natural log of 8

$\ln (e^{\ln (x)}) + \ln (e^{\ln (x^{2})}) = 2 \ln (8)$

Step 1

Simplify the left side.

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

$\ln (e^{\ln (x)} e^{\ln (x^{2})}) = 2 \ln (8)$

Rewrite $\ln (e^{\ln (x)} e^{\ln (x^{2})})$ as $\ln (e^{\ln (x)}) + \ln (e^{\ln (x^{2})})$ .

$\ln (e^{\ln (x)}) + \ln (e^{\ln (x^{2})}) = 2 \ln (8)$

Use logarithm rules to move $\ln (x)$ out of the exponent.

$\ln (x) \ln (e) + \ln (e^{\ln (x^{2})}) = 2 \ln (8)$

The natural logarithm of $e$ is $1$ .

$\ln (x) \cdot 1 + \ln (e^{\ln (x^{2})}) = 2 \ln (8)$

Multiply $\ln (x)$ by $1$ .

$\ln (x) + \ln (e^{\ln (x^{2})}) = 2 \ln (8)$

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

$\ln (x e^{\ln (x^{2})}) = 2 \ln (8)$

Rewrite $\ln (x e^{\ln (x^{2})})$ as $\ln (x) + \ln (e^{\ln (x^{2})})$ .

$\ln (x) + \ln (e^{\ln (x^{2})}) = 2 \ln (8)$

Use logarithm rules to move $\ln (x^{2})$ out of the exponent.

$\ln (x) + \ln (x^{2}) \ln (e) = 2 \ln (8)$

The natural logarithm of $e$ is $1$ .

$\ln (x) + \ln (x^{2}) \cdot 1 = 2 \ln (8)$

Multiply $\ln (x^{2})$ by $1$ .

$\ln (x) + \ln (x^{2}) = 2 \ln (8)$

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

$\ln (x \cdot x^{2}) = 2 \ln (8)$

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

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

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

$\ln (x \cdot x^{2}) = 2 \ln (8)$

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

$\ln (x^{1 + 2}) = 2 \ln (8)$

Add $1$ and $2$ .

$\ln (x^{3}) = 2 \ln (8)$

Step 2

Simplify $2 \ln (8)$ by moving $2$ inside the logarithm.

$\ln (x^{3}) = \ln (8^{2})$

Step 3

For the equation to be equal, the argument of the logarithms on both sides of the equation must be equal.

$x^{3} = 8^{2}$

Step 4

Solve for $x$ .

Subtract $8^{2}$ from both sides of the equation.

$x^{3} - 8^{2} = 0$

Simplify each term.

Raise $8$ to the power of $2$ .

$x^{3} - 1 \cdot 64 = 0$

Multiply $- 1$ by $64$ .

$x^{3} - 64 = 0$

Factor the left side of the equation.

Rewrite $64$ as $4^{3}$ .

$x^{3} - 4^{3} = 0$

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 $b = 4$ .

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

Simplify.

Move $4$ to the left of $x$ .

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

Raise $4$ to the power of $2$ .

$(x - 4) (x^{2} + 4 x + 16) = 0$

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

$x - 4 = 0$

$x^{2} + 4 x + 16 = 0$

Set $x - 4$ equal to $0$ and solve for $x$ .

Set $x - 4$ equal to $0$ .

$x - 4 = 0$

Add $4$ to both sides of the equation.

$x = 4$

Set $x^{2} + 4 x + 16$ equal to $0$ and solve for $x$ .

Set $x^{2} + 4 x + 16$ equal to $0$ .

$x^{2} + 4 x + 16 = 0$

Solve $x^{2} + 4 x + 16 = 0$ for $x$ .

Use the quadratic formula to find the solutions.

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

Substitute the values $a = 1$ , $b = 4$ , and $c = 16$ into the quadratic formula and solve for $x$ .

$\frac{- 4 \pm \sqrt{4^{2} - 4 \cdot (1 \cdot 16)}}{2 \cdot 1}$

Simplify.

Simplify the numerator.

Raise $4$ to the power of $2$ .

$x = \frac{- 4 \pm \sqrt{16 - 4 \cdot 1 \cdot 16}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot 16$ .

Multiply $- 4$ by $1$ .

$x = \frac{- 4 \pm \sqrt{16 - 4 \cdot 16}}{2 \cdot 1}$

Multiply $- 4$ by $16$ .

$x = \frac{- 4 \pm \sqrt{16 - 64}}{2 \cdot 1}$

Subtract $64$ from $16$ .

$x = \frac{- 4 \pm \sqrt{- 48}}{2 \cdot 1}$

Rewrite $- 48$ as $- 1 (48)$ .

$x = \frac{- 4 \pm \sqrt{- 1 \cdot 48}}{2 \cdot 1}$

Rewrite $\sqrt{- 1 (48)}$ as $\sqrt{- 1} \cdot \sqrt{48}$ .

$x = \frac{- 4 \pm \sqrt{- 1} \cdot \sqrt{48}}{2 \cdot 1}$

Rewrite $\sqrt{- 1}$ as $i$ .

$x = \frac{- 4 \pm i \cdot \sqrt{48}}{2 \cdot 1}$

Rewrite $48$ as $4^{2} \cdot 3$ .

Factor $16$ out of $48$ .

$x = \frac{- 4 \pm i \cdot \sqrt{16 (3)}}{2 \cdot 1}$

Rewrite $16$ as $4^{2}$ .

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

Pull terms out from under the radical.

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

Move $4$ to the left of $i$ .

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

Multiply $2$ by $1$ .

$x = \frac{- 4 \pm 4 i \sqrt{3}}{2}$

Simplify $\frac{- 4 \pm 4 i \sqrt{3}}{2}$ .

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

The final answer is the combination of both solutions.

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

The final solution is all the values that make $(x - 4) (x^{2} + 4 x + 16) = 0$ true.

$x = 4, - 2 + 2 i \sqrt{3}, - 2 - 2 i \sqrt{3}$

Solve for x natural log of e^( natural log of x)+ natural log of e^( natural log of x^2)=2 natural log of 8

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