Solve for x (n^(3x-9))^(x+2)=(n^3)^6

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

Step 1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

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

Step 2

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$n^{(3 x - 9) (x + 2)} = n^{3 \cdot 6}$

Step 3

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

$(3 x - 9) (x + 2) = 3 \cdot 6$

Step 4

Solve for $x$ .

Simplify $(3 x - 9) (x + 2)$ .

Expand $(3 x - 9) (x + 2)$ using the FOIL Method.

Apply the distributive property.

$3 x (x + 2) - 9 (x + 2) = 3 \cdot 6$

Apply the distributive property.

$3 x \cdot x + 3 x \cdot 2 - 9 (x + 2) = 3 \cdot 6$

Apply the distributive property.

$3 x \cdot x + 3 x \cdot 2 - 9 x - 9 \cdot 2 = 3 \cdot 6$

Simplify and combine like terms.

Simplify each term.

Multiply $x$ by $x$ by adding the exponents.

Move $x$ .

$3 (x \cdot x) + 3 x \cdot 2 - 9 x - 9 \cdot 2 = 3 \cdot 6$

Multiply $x$ by $x$ .

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

Multiply $2$ by $3$ .

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

Multiply $- 9$ by $2$ .

$3 x^{2} + 6 x - 9 x - 18 = 3 \cdot 6$

Subtract $9 x$ from $6 x$ .

$3 x^{2} - 3 x - 18 = 3 \cdot 6$

Multiply $3$ by $6$ .

$3 x^{2} - 3 x - 18 = 18$

Subtract $18$ from both sides of the equation.

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

Subtract $18$ from $- 18$ .

$3 x^{2} - 3 x - 36 = 0$

Factor the left side of the equation.

Factor $3$ out of $3 x^{2} - 3 x - 36$ .

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

$3 (x^{2}) - 3 x - 36 = 0$

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

$3 (x^{2}) + 3 (- x) - 36 = 0$

Factor $3$ out of $- 36$ .

$3 (x^{2}) + 3 (- x) + 3 (- 12) = 0$

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

$3 (x^{2} - x) + 3 (- 12) = 0$

Factor $3$ out of $3 (x^{2} - x) + 3 (- 12)$ .

$3 (x^{2} - x - 12) = 0$

Factor.

Factor $x^{2} - x - 12$ 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 $- 12$ and whose sum is $- 1$ .

$- 4, 3$

Write the factored form using these integers.

$3 ((x - 4) (x + 3)) = 0$

Remove unnecessary parentheses.

$3 (x - 4) (x + 3) = 0$

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

$x - 4 = 0$

$x + 3 = 0$

Set $x - 4$ equal to $0$ and solve for $x$ .

Set $x - 4$ equal to $0$ .

$x - 4 = 0$

Add $4$ to both sides of the equation.

$x = 4$

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.

$x = - 3$

The final solution is all the values that make $3 (x - 4) (x + 3) = 0$ true.

$x = 4, - 3$

Solve for x (n^(3x-9))^(x+2)=(n^3)^6

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