Precalculus Examples

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

Step 1

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

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

Step 2

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

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

Step 3

Solve for $x$ .

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Step 3.1

Subtract $3$ from both sides of the equation.

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

Step 3.2

Factor $x^{2} - 2 x - 3$ using the AC method.

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Step 3.2.1

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 $- 3$ and whose sum is $- 2$ .

$- 3, 1$

Step 3.2.2

Write the factored form using these integers.

$(x - 3) (x + 1) = 0$

Step 3.3

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

$x - 3 = 0$

$x + 1 = 0$

Step 3.4

Set $x - 3$ equal to $0$ and solve for $x$ .

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Step 3.4.1

Set $x - 3$ equal to $0$ .

$x - 3 = 0$

Step 3.4.2

Add $3$ to both sides of the equation.

$x = 3$

Step 3.5

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

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Step 3.5.1

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

$x + 1 = 0$

Step 3.5.2

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 3.6

The final solution is all the values that make $(x - 3) (x + 1) = 0$ true.

$x = 3, - 1$