Algebra Examples

${(\frac{1}{25})}^{- 16 x - 62} = 5^{- 2 x^{2} - 2}$

Step 1

Apply the product rule to $\frac{1}{25}$ .

$\frac{1^{- 16 x - 62}}{25^{- 16 x - 62}} = 5^{- 2 x^{2} - 2}$

Step 2

One to any power is one.

$\frac{1}{25^{- 16 x - 62}} = 5^{- 2 x^{2} - 2}$

Step 3

Move $25^{- 16 x - 62}$ to the numerator using the negative exponent rule $\frac{1}{b^{- n}} = b^{n}$ .

$25^{- (- 16 x - 62)} = 5^{- 2 x^{2} - 2}$

Step 4

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

$5^{2 (- (- 16 x - 62))} = 5^{- 2 x^{2} - 2}$

Step 5

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

$2 (- (- 16 x - 62)) = - 2 x^{2} - 2$

Step 6

Solve for $x$ .

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Step 6.1

Simplify $2 (- (- 16 x - 62))$ .

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Step 6.1.1

Rewrite.

$0 + 0 + 2 (- (- 16 x - 62)) = - 2 x^{2} - 2$

Step 6.1.2

Simplify by adding zeros.

$2 (- (- 16 x - 62)) = - 2 x^{2} - 2$

Step 6.1.3

Apply the distributive property.

$2 (- (- 16 x) - - 62) = - 2 x^{2} - 2$

Step 6.1.4

Multiply.

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Step 6.1.4.1

Multiply $- 16$ by $- 1$ .

$2 (16 x - - 62) = - 2 x^{2} - 2$

Step 6.1.4.2

Multiply $- 1$ by $- 62$ .

$2 (16 x + 62) = - 2 x^{2} - 2$

Step 6.1.5

Apply the distributive property.

$2 (16 x) + 2 \cdot 62 = - 2 x^{2} - 2$

Step 6.1.6

Multiply.

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Step 6.1.6.1

Multiply $16$ by $2$ .

$32 x + 2 \cdot 62 = - 2 x^{2} - 2$

Step 6.1.6.2

Multiply $2$ by $62$ .

$32 x + 124 = - 2 x^{2} - 2$

Step 6.2

Add $2 x^{2}$ to both sides of the equation.

$32 x + 124 + 2 x^{2} = - 2$

Step 6.3

Add $2$ to both sides of the equation.

$32 x + 124 + 2 x^{2} + 2 = 0$

Step 6.4

Add $124$ and $2$ .

$32 x + 2 x^{2} + 126 = 0$

Step 6.5

Factor the left side of the equation.

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Step 6.5.1

Factor $2$ out of $32 x + 2 x^{2} + 126$ .

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Step 6.5.1.1

Factor $2$ out of $32 x$ .

$2 (16 x) + 2 x^{2} + 126 = 0$

Step 6.5.1.2

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

$2 (16 x) + 2 (x^{2}) + 126 = 0$

Step 6.5.1.3

Factor $2$ out of $126$ .

$2 (16 x) + 2 x^{2} + 2 \cdot 63 = 0$

Step 6.5.1.4

Factor $2$ out of $2 (16 x) + 2 x^{2}$ .

$2 (16 x + x^{2}) + 2 \cdot 63 = 0$

Step 6.5.1.5

Factor $2$ out of $2 (16 x + x^{2}) + 2 \cdot 63$ .

$2 (16 x + x^{2} + 63) = 0$

Step 6.5.2

Let $u = x$ . Substitute $u$ for all occurrences of $x$ .

$2 (16 u + u^{2} + 63) = 0$

Step 6.5.3

Factor $16 u + u^{2} + 63$ using the AC method.

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Step 6.5.3.1

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 $63$ and whose sum is $16$ .

$7, 9$

Step 6.5.3.2

Write the factored form using these integers.

$2 ((u + 7) (u + 9)) = 0$

Step 6.5.4

Factor.

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Step 6.5.4.1

Replace all occurrences of $u$ with $x$ .

$2 ((x + 7) (x + 9)) = 0$

Step 6.5.4.2

Remove unnecessary parentheses.

$2 (x + 7) (x + 9) = 0$

Step 6.6

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

$x + 7 = 0$

$x + 9 = 0$

Step 6.7

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

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Step 6.7.1

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

$x + 7 = 0$

Step 6.7.2

Subtract $7$ from both sides of the equation.

$x = - 7$

Step 6.8

Set $x + 9$ equal to $0$ and solve for $x$ .

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Step 6.8.1

Set $x + 9$ equal to $0$ .

$x + 9 = 0$

Step 6.8.2

Subtract $9$ from both sides of the equation.

$x = - 9$

Step 6.9

The final solution is all the values that make $2 (x + 7) (x + 9) = 0$ true.

$x = - 7, - 9$