Precalculus Examples

$\frac{2^{x^{2}}}{64^{x}} = 128$

Step 1

Move $64^{x}$ to the numerator using the negative exponent rule $\frac{1}{b^{- n}} = b^{n}$ .

$2^{x^{2}} \cdot 64^{- x} = 128$

Step 2

Rewrite $64$ as $2^{6}$ .

$2^{x^{2}} \cdot {(2^{6})}^{- x} = 128$

Step 3

Multiply the exponents in ${(2^{6})}^{- x}$ .

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Step 3.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$2^{x^{2}} \cdot 2^{6 (- x)} = 128$

Step 3.2

Multiply $- 1$ by $6$ .

$2^{x^{2}} \cdot 2^{- 6 x} = 128$

Step 4

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$2^{x^{2} - 6 x} = 128$

Step 5

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

$2^{x^{2} - 6 x} = 2^{7}$

Step 6

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

$x^{2} - 6 x = 7$

Step 7

Solve for $x$ .

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Step 7.1

Subtract $7$ from both sides of the equation.

$x^{2} - 6 x - 7 = 0$

Step 7.2

Factor $x^{2} - 6 x - 7$ using the AC method.

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Step 7.2.1

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 $- 7$ and whose sum is $- 6$ .

$- 7, 1$

Step 7.2.2

Write the factored form using these integers.

$(x - 7) (x + 1) = 0$

Step 7.3

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

$x - 7 = 0$

$x + 1 = 0$

Step 7.4

Set $x - 7$ equal to $0$ and solve for $x$ .

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Step 7.4.1

Set $x - 7$ equal to $0$ .

$x - 7 = 0$

Step 7.4.2

Add $7$ to both sides of the equation.

$x = 7$

Step 7.5

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

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Step 7.5.1

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

$x + 1 = 0$

Step 7.5.2

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 7.6

The final solution is all the values that make $(x - 7) (x + 1) = 0$ true.

$x = 7, - 1$