Solve for x (1/7)^(3x^2+3x)=49^(-6x+3)

${(\frac{1}{7})}^{3 x^{2} + 3 x} = 49^{- 6 x + 3}$

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

$\frac{1^{3 x^{2} + 3 x}}{7^{3 x^{2} + 3 x}} = 49^{- 6 x + 3}$

One to any power is one.

$\frac{1}{7^{3 x^{2} + 3 x}} = 49^{- 6 x + 3}$

Move $7^{3 x^{2} + 3 x}$ to the numerator using the negative exponent rule $\frac{1}{b^{- n}} = b^{n}$ .

$7^{- (3 x^{2} + 3 x)} = 49^{- 6 x + 3}$

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

$7^{- (3 x^{2} + 3 x)} = 7^{2 (- 6 x + 3)}$

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

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

Solve for $x$ .

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

Start simplifying.

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

Simplify by adding zeros.

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

Apply the distributive property.

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

Multiply.

Multiply $3$ by $- 1$ .

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

Multiply $3$ by $- 1$ .

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

Simplify $2 (- 6 x + 3)$ .

Apply the distributive property.

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

Multiply.

Multiply $- 6$ by $2$ .

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

Multiply $2$ by $3$ .

$- 3 x^{2} - 3 x = - 12 x + 6$

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

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

$- 3 x^{2} - 3 x + 12 x = 6$

Add $- 3 x$ and $12 x$ .

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

Subtract $6$ from both sides of the equation.

$- 3 x^{2} + 9 x - 6 = 0$

Factor the left side of the equation.

Factor $- 3$ out of $- 3 x^{2} + 9 x - 6$ .

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

$- 3 x^{2} + 9 x - 6 = 0$

Factor $- 3$ out of $9 x$ .

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

Factor $- 3$ out of $- 6$ .

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

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

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

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

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

Factor.

Factor $x^{2} - 3 x + 2$ 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 $2$ and whose sum is $- 3$ .

$- 2, - 1$

Write the factored form using these integers.

$- 3 ((x - 2) (x - 1)) = 0$

Remove unnecessary parentheses.

$- 3 (x - 2) (x - 1) = 0$

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

$x - 2 = 0$

$x - 1 = 0$

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

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

$x - 2 = 0$

Add $2$ to both sides of the equation.

$x = 2$

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

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

$x - 1 = 0$

Add $1$ to both sides of the equation.

$x = 1$

The final solution is all the values that make $- 3 (x - 2) (x - 1) = 0$ true.

$x = 2, 1$

Solve for x (1/7)^(3x^2+3x)=49^(-6x+3)

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