Precalculus Examples

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

Step 1

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

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

Step 2

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

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

Step 3

Solve for $x$ .

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Step 3.1

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

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

Step 3.2

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

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

Step 3.3

Use the quadratic formula to find the solutions.

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

Step 3.4

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

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

Step 3.5

Simplify.

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Step 3.5.1

Simplify the numerator.

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Step 3.5.1.1

Multiply $- 4 \cdot 2 \cdot - 1$ .

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Step 3.5.1.1.1

Multiply $- 4$ by $2$ .

$x = \frac{- e \pm \sqrt{e^{2} - 8 \cdot (- 1 e^{2})}}{2 \cdot 2}$

Step 3.5.1.1.2

Multiply $- 8$ by $- 1$ .

$x = \frac{- e \pm \sqrt{e^{2} + 8 e^{2}}}{2 \cdot 2}$

Step 3.5.1.2

Add $e^{2}$ and $8 e^{2}$ .

$x = \frac{- e \pm \sqrt{9 e^{2}}}{2 \cdot 2}$

Step 3.5.1.3

Rewrite $9 e^{2}$ as ${(3 e)}^{2}$ .

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

Step 3.5.1.4

Pull terms out from under the radical, assuming positive real numbers.

$x = \frac{- e \pm 3 e}{2 \cdot 2}$

Step 3.5.2

Multiply $2$ by $2$ .

$x = \frac{- e \pm 3 e}{4}$

Step 3.6

The final answer is the combination of both solutions.

$x = \frac{e}{2}, - e$

Step 4

The result can be shown in multiple forms.

Exact Form:

$x = \frac{e}{2}, - e$

Decimal Form:

$x = 1.35914091 \dots, - 2.71828182 \dots$