Solve for x 5^(-3x^2+15)=(1/25)^(-9x+33)

$5^{- 3 x^{2} + 15} = {(\frac{1}{25})}^{- 9 x + 33}$

Step 1

Apply the product rule to $\frac{1}{25}$ .

$5^{- 3 x^{2} + 15} = \frac{1^{- 9 x + 33}}{25^{- 9 x + 33}}$

Step 2

One to any power is one.

$5^{- 3 x^{2} + 15} = \frac{1}{25^{- 9 x + 33}}$

Step 3

Move $25^{- 9 x + 33}$ to the numerator using the negative exponent rule $\frac{1}{b^{- n}} = b^{n}$ .

$5^{- 3 x^{2} + 15} = 25^{- (- 9 x + 33)}$

Step 4

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

$5^{- 3 x^{2} + 15} = 5^{2 (- (- 9 x + 33))}$

Step 5

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

$- 3 x^{2} + 15 = 2 (- (- 9 x + 33))$

Step 6

Solve for $x$ .

Simplify $2 (- (- 9 x + 33))$ .

Rewrite.

$- 3 x^{2} + 15 = 0 + 0 + 2 (- (- 9 x + 33))$

Simplify by adding zeros.

$- 3 x^{2} + 15 = 2 (- (- 9 x + 33))$

Apply the distributive property.

$- 3 x^{2} + 15 = 2 (- (- 9 x) - 1 \cdot 33)$

Multiply.

Multiply $- 9$ by $- 1$ .

$- 3 x^{2} + 15 = 2 (9 x - 1 \cdot 33)$

Multiply $- 1$ by $33$ .

$- 3 x^{2} + 15 = 2 (9 x - 33)$

Apply the distributive property.

$- 3 x^{2} + 15 = 2 (9 x) + 2 \cdot - 33$

Multiply.

Multiply $9$ by $2$ .

$- 3 x^{2} + 15 = 18 x + 2 \cdot - 33$

Multiply $2$ by $- 33$ .

$- 3 x^{2} + 15 = 18 x - 66$

Subtract $18 x$ from both sides of the equation.

$- 3 x^{2} + 15 - 18 x = - 66$

Add $66$ to both sides of the equation.

$- 3 x^{2} + 15 - 18 x + 66 = 0$

Add $15$ and $66$ .

$- 3 x^{2} - 18 x + 81 = 0$

Factor the left side of the equation.

Factor $- 3$ out of $- 3 x^{2} - 18 x + 81$ .

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

$- 3 x^{2} - 18 x + 81 = 0$

Factor $- 3$ out of $- 18 x$ .

$- 3 x^{2} - 3 (6 x) + 81 = 0$

Factor $- 3$ out of $81$ .

$- 3 x^{2} - 3 (6 x) - 3 \cdot - 27 = 0$

Factor $- 3$ out of $- 3 (x^{2}) - 3 (6 x)$ .

$- 3 (x^{2} + 6 x) - 3 \cdot - 27 = 0$

Factor $- 3$ out of $- 3 (x^{2} + 6 x) - 3 (- 27)$ .

$- 3 (x^{2} + 6 x - 27) = 0$

Factor.

Factor $x^{2} + 6 x - 27$ 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 $- 27$ and whose sum is $6$ .

$- 3, 9$

Write the factored form using these integers.

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

Remove unnecessary parentheses.

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

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

$x - 3 = 0$

$x + 9 = 0$

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

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

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

Set $x + 9$ equal to $0$ and solve for $x$ .

Set $x + 9$ equal to $0$ .

$x + 9 = 0$

Subtract $9$ from both sides of the equation.

$x = - 9$

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

$x = 3, - 9$

Solve for x 5^(-3x^2+15)=(1/25)^(-9x+33)

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