Evaluate e^x=e^(x^2-2)

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

Step 1

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

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

Step 2

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

$x = x^{2} - 2$

Step 3

Solve for $x$ .

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

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

Add $2$ to both sides of the equation.

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

Factor the left side of the equation.

Factor $- 1$ out of $x - x^{2} + 2$ .

Reorder $x$ and $- x^{2}$ .

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

Factor $- 1$ out of $- x^{2}$ .

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

Factor $- 1$ out of $x$ .

$- (x^{2}) - 1 (- x) + 2 = 0$

Rewrite $2$ as $- 1 (- 2)$ .

$- (x^{2}) - 1 (- x) - 1 \cdot - 2 = 0$

Factor $- 1$ out of $- (x^{2}) - 1 (- x)$ .

$- (x^{2} - x) - 1 \cdot - 2 = 0$

Factor $- 1$ out of $- (x^{2} - x) - 1 (- 2)$ .

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

Factor.

Factor $x^{2} - x - 2$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 2$ and whose sum is $- 1$ .

$- 2, 1$

Write the factored form using these integers.

$- ((x - 2) (x + 1)) = 0$

Remove unnecessary parentheses.

$- (x - 2) (x + 1) = 0$

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

$x - 2 = 0$

$x + 1 = 0$

Set $x - 2$ equal to $0$ and solve for $x$ .

Set $x - 2$ equal to $0$ .

$x - 2 = 0$

Add $2$ to both sides of the equation.

$x = 2$

Set $x + 1$ equal to $0$ and solve for $x$ .

Set $x + 1$ equal to $0$ .

$x + 1 = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$

The final solution is all the values that make $- (x - 2) (x + 1) = 0$ true.

$x = 2, - 1$

Evaluate e^x=e^(x^2-2)

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