Algebra Examples

$25^{- 6 x - 69} = {(\frac{1}{5})}^{3 x^{2} + 3}$

Step 1

Apply the product rule to $\frac{1}{5}$ .

$25^{- 6 x - 69} = \frac{1^{3 x^{2} + 3}}{5^{3 x^{2} + 3}}$

Step 2

One to any power is one.

$25^{- 6 x - 69} = \frac{1}{5^{3 x^{2} + 3}}$

Step 3

Move $5^{3 x^{2} + 3}$ to the numerator using the negative exponent rule $\frac{1}{b^{- n}} = b^{n}$ .

$25^{- 6 x - 69} = 5^{- (3 x^{2} + 3)}$

Step 4

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

$5^{2 (- 6 x - 69)} = 5^{- (3 x^{2} + 3)}$

Step 5

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

$2 (- 6 x - 69) = - (3 x^{2} + 3)$

Step 6

Solve for $x$ .

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Step 6.1

Simplify $2 (- 6 x - 69)$ .

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Step 6.1.1

Rewrite.

$0 + 0 + 2 (- 6 x - 69) = - (3 x^{2} + 3)$

Step 6.1.2

Simplify by adding zeros.

$2 (- 6 x - 69) = - (3 x^{2} + 3)$

Step 6.1.3

Apply the distributive property.

$2 (- 6 x) + 2 \cdot - 69 = - (3 x^{2} + 3)$

Step 6.1.4

Multiply.

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Step 6.1.4.1

Multiply $- 6$ by $2$ .

$- 12 x + 2 \cdot - 69 = - (3 x^{2} + 3)$

Step 6.1.4.2

Multiply $2$ by $- 69$ .

$- 12 x - 138 = - (3 x^{2} + 3)$

Step 6.2

Simplify $- (3 x^{2} + 3)$ .

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Step 6.2.1

Apply the distributive property.

$- 12 x - 138 = - (3 x^{2}) - 1 \cdot 3$

Step 6.2.2

Multiply.

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Step 6.2.2.1

Multiply $3$ by $- 1$ .

$- 12 x - 138 = - 3 x^{2} - 1 \cdot 3$

Step 6.2.2.2

Multiply $- 1$ by $3$ .

$- 12 x - 138 = - 3 x^{2} - 3$

Step 6.3

Add $3 x^{2}$ to both sides of the equation.

$- 12 x - 138 + 3 x^{2} = - 3$

Step 6.4

Add $3$ to both sides of the equation.

$- 12 x - 138 + 3 x^{2} + 3 = 0$

Step 6.5

Add $- 138$ and $3$ .

$- 12 x + 3 x^{2} - 135 = 0$

Step 6.6

Factor the left side of the equation.

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Step 6.6.1

Factor $3$ out of $- 12 x + 3 x^{2} - 135$ .

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Step 6.6.1.1

Factor $3$ out of $- 12 x$ .

$3 (- 4 x) + 3 x^{2} - 135 = 0$

Step 6.6.1.2

Factor $3$ out of $3 x^{2}$ .

$3 (- 4 x) + 3 (x^{2}) - 135 = 0$

Step 6.6.1.3

Factor $3$ out of $- 135$ .

$3 (- 4 x) + 3 x^{2} + 3 \cdot - 45 = 0$

Step 6.6.1.4

Factor $3$ out of $3 (- 4 x) + 3 x^{2}$ .

$3 (- 4 x + x^{2}) + 3 \cdot - 45 = 0$

Step 6.6.1.5

Factor $3$ out of $3 (- 4 x + x^{2}) + 3 \cdot - 45$ .

$3 (- 4 x + x^{2} - 45) = 0$

Step 6.6.2

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

$3 (- 4 u + u^{2} - 45) = 0$

Step 6.6.3

Factor $- 4 u + u^{2} - 45$ using the AC method.

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Step 6.6.3.1

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 $- 45$ and whose sum is $- 4$ .

$- 9, 5$

Step 6.6.3.2

Write the factored form using these integers.

$3 ((u - 9) (u + 5)) = 0$

Step 6.6.4

Factor.

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Step 6.6.4.1

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

$3 ((x - 9) (x + 5)) = 0$

Step 6.6.4.2

Remove unnecessary parentheses.

$3 (x - 9) (x + 5) = 0$

Step 6.7

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

$x - 9 = 0$

$x + 5 = 0$

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

Add $9$ to both sides of the equation.

$x = 9$

Step 6.9

Set $x + 5$ equal to $0$ and solve for $x$ .

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Step 6.9.1

Set $x + 5$ equal to $0$ .

$x + 5 = 0$

Step 6.9.2

Subtract $5$ from both sides of the equation.

$x = - 5$

Step 6.10

The final solution is all the values that make $3 (x - 9) (x + 5) = 0$ true.

$x = 9, - 5$