Solve for x (1/7)^(-2x^2-8x)=343^(-4x-14)

${(\frac{1}{7})}^{- 2 x^{2} - 8 x} = 343^{- 4 x - 14}$

Step 1

Apply the product rule to $\frac{1}{7}$ .

$\frac{1^{- 2 x^{2} - 8 x}}{7^{- 2 x^{2} - 8 x}} = 343^{- 4 x - 14}$

Step 2

One to any power is one.

$\frac{1}{7^{- 2 x^{2} - 8 x}} = 343^{- 4 x - 14}$

Step 3

Move $7^{- 2 x^{2} - 8 x}$ to the numerator using the negative exponent rule $\frac{1}{b^{- n}} = b^{n}$ .

$7^{- (- 2 x^{2} - 8 x)} = 343^{- 4 x - 14}$

Step 4

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

$7^{- (- 2 x^{2} - 8 x)} = 7^{3 (- 4 x - 14)}$

Step 5

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

$- (- 2 x^{2} - 8 x) = 3 (- 4 x - 14)$

Step 6

Solve for $x$ .

Simplify $- (- 2 x^{2} - 8 x)$ .

Start simplifying.

$0 + 0 - (- 2 x^{2} - 8 x) = 3 (- 4 x - 14)$

Simplify by adding zeros.

$- (- 2 x^{2} - 8 x) = 3 (- 4 x - 14)$

Apply the distributive property.

$- (- 2 x^{2}) - (- 8 x) = 3 (- 4 x - 14)$

Multiply.

Multiply $- 2$ by $- 1$ .

$2 x^{2} - (- 8 x) = 3 (- 4 x - 14)$

Multiply $- 8$ by $- 1$ .

$2 x^{2} + 8 x = 3 (- 4 x - 14)$

Simplify $3 (- 4 x - 14)$ .

Apply the distributive property.

$2 x^{2} + 8 x = 3 (- 4 x) + 3 \cdot - 14$

Multiply.

Multiply $- 4$ by $3$ .

$2 x^{2} + 8 x = - 12 x + 3 \cdot - 14$

Multiply $3$ by $- 14$ .

$2 x^{2} + 8 x = - 12 x - 42$

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

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

$2 x^{2} + 8 x + 12 x = - 42$

Add $8 x$ and $12 x$ .

$2 x^{2} + 20 x = - 42$

Add $42$ to both sides of the equation.

$2 x^{2} + 20 x + 42 = 0$

Factor the left side of the equation.

Factor $2$ out of $2 x^{2} + 20 x + 42$ .

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

$2 (x^{2}) + 20 x + 42 = 0$

Factor $2$ out of $20 x$ .

$2 (x^{2}) + 2 (10 x) + 42 = 0$

Factor $2$ out of $42$ .

$2 x^{2} + 2 (10 x) + 2 \cdot 21 = 0$

Factor $2$ out of $2 x^{2} + 2 (10 x)$ .

$2 (x^{2} + 10 x) + 2 \cdot 21 = 0$

Factor $2$ out of $2 (x^{2} + 10 x) + 2 \cdot 21$ .

$2 (x^{2} + 10 x + 21) = 0$

Factor.

Factor $x^{2} + 10 x + 21$ 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 $21$ and whose sum is $10$ .

$3, 7$

Write the factored form using these integers.

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

Remove unnecessary parentheses.

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

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

$x + 3 = 0$

$x + 7 = 0$

Set $x + 3$ equal to $0$ and solve for $x$ .

Set $x + 3$ equal to $0$ .

$x + 3 = 0$

Subtract $3$ from both sides of the equation.

$x = - 3$

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

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

$x + 7 = 0$

Subtract $7$ from both sides of the equation.

$x = - 7$

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

$x = - 3, - 7$

Solve for x (1/7)^(-2x^2-8x)=343^(-4x-14)

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