Precalculus Examples

$e^{x^{2} - 3} = e^{x - 2}$

Step 1

Create equivalent expressions in the equation that all have equal bases.

$e^{x^{2} - 3} = e^{x - 2}$

Step 2

Since the bases are the same, then two expressions are only equal if the exponents are also equal.

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

Step 3

Solve for $x$ .

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Step 3.1

Subtract $x$ from both sides of the equation.

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

Step 3.2

Move all terms to the left side of the equation and simplify.

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Step 3.2.1

Add $2$ to both sides of the equation.

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

Step 3.2.2

Add $- 3$ and $2$ .

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

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

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

Step 3.5

Simplify.

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Step 3.5.1

Simplify the numerator.

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Step 3.5.1.1

Raise $- 1$ to the power of $2$ .

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

Step 3.5.1.2

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

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Step 3.5.1.2.1

Multiply $- 4$ by $1$ .

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

Step 3.5.1.2.2

Multiply $- 4$ by $- 1$ .

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

Step 3.5.1.3

Add $1$ and $4$ .

$x = \frac{1 \pm \sqrt{5}}{2 \cdot 1}$

Step 3.5.2

Multiply $2$ by $1$ .

$x = \frac{1 \pm \sqrt{5}}{2}$

Step 3.6

The final answer is the combination of both solutions.

$x = \frac{1 + \sqrt{5}}{2}, \frac{1 - \sqrt{5}}{2}$

Step 4

The result can be shown in multiple forms.

Exact Form:

$x = \frac{1 + \sqrt{5}}{2}, \frac{1 - \sqrt{5}}{2}$

Decimal Form:

$x = 1.61803398 \dots, - 0.61803398 \dots$