Precalculus Examples

$2^{x^{2} + 5 x} = \frac{1}{16}$

Step 1

Raise $16$ to the power of $1$ .

$2^{x^{2} + 5 x} = \frac{1}{16^{1}}$

Step 2

Move $16^{1}$ to the numerator using the negative exponent rule $\frac{1}{b^{n}} = b^{- n}$ .

$2^{x^{2} + 5 x} = 16^{- 1}$

Step 3

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

$2^{x^{2} + 5 x} = 2^{- 4}$

Step 4

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

$x^{2} + 5 x = - 4$

Step 5

Solve for $x$ .

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Step 5.1

Add $4$ to both sides of the equation.

$x^{2} + 5 x + 4 = 0$

Step 5.2

Factor $x^{2} + 5 x + 4$ using the AC method.

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Step 5.2.1

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 $4$ and whose sum is $5$ .

$1, 4$

Step 5.2.2

Write the factored form using these integers.

$(x + 1) (x + 4) = 0$

Step 5.3

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

$x + 1 = 0$

$x + 4 = 0$

Step 5.4

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

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Step 5.4.1

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

$x + 1 = 0$

Step 5.4.2

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 5.5

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

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Step 5.5.1

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

$x + 4 = 0$

Step 5.5.2

Subtract $4$ from both sides of the equation.

$x = - 4$

Step 5.6

The final solution is all the values that make $(x + 1) (x + 4) = 0$ true.

$x = - 1, - 4$