Solve for x 3^(-3x^2+3x)=(1/9)^(6x+21)

$3^{- 3 x^{2} + 3 x} = {(\frac{1}{9})}^{6 x + 21}$

Step 1

Apply the product rule to $\frac{1}{9}$ .

$3^{- 3 x^{2} + 3 x} = \frac{1^{6 x + 21}}{9^{6 x + 21}}$

Step 2

One to any power is one.

$3^{- 3 x^{2} + 3 x} = \frac{1}{9^{6 x + 21}}$

Step 3

Move $9^{6 x + 21}$ to the numerator using the negative exponent rule $\frac{1}{b^{- n}} = b^{n}$ .

$3^{- 3 x^{2} + 3 x} = 9^{- (6 x + 21)}$

Step 4

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

$3^{- 3 x^{2} + 3 x} = 3^{2 (- (6 x + 21))}$

Step 5

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

$- 3 x^{2} + 3 x = 2 (- (6 x + 21))$

Step 6

Solve for $x$ .

Simplify $2 (- (6 x + 21))$ .

Apply the distributive property.

$- 3 x^{2} + 3 x = 2 (- (6 x) - 1 \cdot 21)$

Multiply.

Multiply $6$ by $- 1$ .

$- 3 x^{2} + 3 x = 2 (- 6 x - 1 \cdot 21)$

Multiply $- 1$ by $21$ .

$- 3 x^{2} + 3 x = 2 (- 6 x - 21)$

Apply the distributive property.

$- 3 x^{2} + 3 x = 2 (- 6 x) + 2 \cdot - 21$

Multiply.

Multiply $- 6$ by $2$ .

$- 3 x^{2} + 3 x = - 12 x + 2 \cdot - 21$

Multiply $2$ by $- 21$ .

$- 3 x^{2} + 3 x = - 12 x - 42$

Move all terms containing $x$ to the left side of the equation.

Add $12 x$ to both sides of the equation.

$- 3 x^{2} + 3 x + 12 x = - 42$

Add $3 x$ and $12 x$ .

$- 3 x^{2} + 15 x = - 42$

Add $42$ to both sides of the equation.

$- 3 x^{2} + 15 x + 42 = 0$

Factor the left side of the equation.

Factor $- 3$ out of $- 3 x^{2} + 15 x + 42$ .

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

$- 3 x^{2} + 15 x + 42 = 0$

Factor $- 3$ out of $15 x$ .

$- 3 x^{2} - 3 (- 5 x) + 42 = 0$

Factor $- 3$ out of $42$ .

$- 3 x^{2} - 3 (- 5 x) - 3 \cdot - 14 = 0$

Factor $- 3$ out of $- 3 (x^{2}) - 3 (- 5 x)$ .

$- 3 (x^{2} - 5 x) - 3 \cdot - 14 = 0$

Factor $- 3$ out of $- 3 (x^{2} - 5 x) - 3 (- 14)$ .

$- 3 (x^{2} - 5 x - 14) = 0$

Factor.

Factor $x^{2} - 5 x - 14$ using the AC method.

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 $- 14$ and whose sum is $- 5$ .

$- 7, 2$

Write the factored form using these integers.

$- 3 ((x - 7) (x + 2)) = 0$

Remove unnecessary parentheses.

$- 3 (x - 7) (x + 2) = 0$

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

$x - 7 = 0$

$x + 2 = 0$

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

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

$x - 7 = 0$

Add $7$ to both sides of the equation.

$x = 7$

Set $x + 2$ equal to $0$ and solve for $x$ .

Set $x + 2$ equal to $0$ .

$x + 2 = 0$

Subtract $2$ from both sides of the equation.

$x = - 2$

The final solution is all the values that make $- 3 (x - 7) (x + 2) = 0$ true.

$x = 7, - 2$

Solve for x 3^(-3x^2+3x)=(1/9)^(6x+21)

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