Solve for x (1/9)^(-2x+46)=3^(-2x^2+4)

${(\frac{1}{9})}^{- 2 x + 46} = 3^{- 2 x^{2} + 4}$

Step 1

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

$\frac{1^{- 2 x + 46}}{9^{- 2 x + 46}} = 3^{- 2 x^{2} + 4}$

Step 2

One to any power is one.

$\frac{1}{9^{- 2 x + 46}} = 3^{- 2 x^{2} + 4}$

Step 3

Move $9^{- 2 x + 46}$ to the numerator using the negative exponent rule $\frac{1}{b^{- n}} = b^{n}$ .

$9^{- (- 2 x + 46)} = 3^{- 2 x^{2} + 4}$

Step 4

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

$3^{2 (- (- 2 x + 46))} = 3^{- 2 x^{2} + 4}$

Step 5

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

$2 (- (- 2 x + 46)) = - 2 x^{2} + 4$

Step 6

Solve for $x$ .

Simplify $2 (- (- 2 x + 46))$ .

Rewrite.

$0 + 0 + 2 (- (- 2 x + 46)) = - 2 x^{2} + 4$

Simplify by adding zeros.

$2 (- (- 2 x + 46)) = - 2 x^{2} + 4$

Apply the distributive property.

$2 (- (- 2 x) - 1 \cdot 46) = - 2 x^{2} + 4$

Multiply.

Multiply $- 2$ by $- 1$ .

$2 (2 x - 1 \cdot 46) = - 2 x^{2} + 4$

Multiply $- 1$ by $46$ .

$2 (2 x - 46) = - 2 x^{2} + 4$

Apply the distributive property.

$2 (2 x) + 2 \cdot - 46 = - 2 x^{2} + 4$

Multiply.

Multiply $2$ by $2$ .

$4 x + 2 \cdot - 46 = - 2 x^{2} + 4$

Multiply $2$ by $- 46$ .

$4 x - 92 = - 2 x^{2} + 4$

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

$4 x - 92 + 2 x^{2} = 4$

Subtract $4$ from both sides of the equation.

$4 x - 92 + 2 x^{2} - 4 = 0$

Subtract $4$ from $- 92$ .

$4 x + 2 x^{2} - 96 = 0$

Factor the left side of the equation.

Factor $2$ out of $4 x + 2 x^{2} - 96$ .

Factor $2$ out of $4 x$ .

$2 (2 x) + 2 x^{2} - 96 = 0$

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

$2 (2 x) + 2 (x^{2}) - 96 = 0$

Factor $2$ out of $- 96$ .

$2 (2 x) + 2 x^{2} + 2 \cdot - 48 = 0$

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

$2 (2 x + x^{2}) + 2 \cdot - 48 = 0$

Factor $2$ out of $2 (2 x + x^{2}) + 2 \cdot - 48$ .

$2 (2 x + x^{2} - 48) = 0$

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

$2 (2 u + u^{2} - 48) = 0$

Factor $2 u + u^{2} - 48$ 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 $- 48$ and whose sum is $2$ .

$- 6, 8$

Write the factored form using these integers.

$2 ((u - 6) (u + 8)) = 0$

Factor.

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

$2 ((x - 6) (x + 8)) = 0$

Remove unnecessary parentheses.

$2 (x - 6) (x + 8) = 0$

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

$x - 6 = 0$

$x + 8 = 0$

Set $x - 6$ equal to $0$ and solve for $x$ .

Set $x - 6$ equal to $0$ .

$x - 6 = 0$

Add $6$ to both sides of the equation.

$x = 6$

Set $x + 8$ equal to $0$ and solve for $x$ .

Set $x + 8$ equal to $0$ .

$x + 8 = 0$

Subtract $8$ from both sides of the equation.

$x = - 8$

The final solution is all the values that make $2 (x - 6) (x + 8) = 0$ true.

$x = 6, - 8$

Solve for x (1/9)^(-2x+46)=3^(-2x^2+4)

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