Solve for x (1/8)^(3x^2-3x)=64^(9x-21)

${(\frac{1}{8})}^{3 x^{2} - 3 x} = 64^{9 x - 21}$

Step 1

Apply the product rule to $\frac{1}{8}$ .

$\frac{1^{3 x^{2} - 3 x}}{8^{3 x^{2} - 3 x}} = 64^{9 x - 21}$

Step 2

One to any power is one.

$\frac{1}{8^{3 x^{2} - 3 x}} = 64^{9 x - 21}$

Step 3

Move $8^{3 x^{2} - 3 x}$ to the numerator using the negative exponent rule $\frac{1}{b^{- n}} = b^{n}$ .

$8^{- (3 x^{2} - 3 x)} = 64^{9 x - 21}$

Step 4

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

$8^{- (3 x^{2} - 3 x)} = 8^{2 (9 x - 21)}$

Step 5

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

$- (3 x^{2} - 3 x) = 2 (9 x - 21)$

Step 6

Solve for $x$ .

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

Rewrite.

$0 + 0 - (3 x^{2} - 3 x) = 2 (9 x - 21)$

Simplify by adding zeros.

$- (3 x^{2} - 3 x) = 2 (9 x - 21)$

Apply the distributive property.

$- (3 x^{2}) - (- 3 x) = 2 (9 x - 21)$

Multiply.

Multiply $3$ by $- 1$ .

$- 3 x^{2} - (- 3 x) = 2 (9 x - 21)$

Multiply $- 3$ by $- 1$ .

$- 3 x^{2} + 3 x = 2 (9 x - 21)$

Simplify $2 (9 x - 21)$ .

Apply the distributive property.

$- 3 x^{2} + 3 x = 2 (9 x) + 2 \cdot - 21$

Multiply.

Multiply $9$ by $2$ .

$- 3 x^{2} + 3 x = 18 x + 2 \cdot - 21$

Multiply $2$ by $- 21$ .

$- 3 x^{2} + 3 x = 18 x - 42$

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

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

$- 3 x^{2} + 3 x - 18 x = - 42$

Subtract $18 x$ from $3 x$ .

$- 3 x^{2} - 15 x = - 42$

Add $42$ to both sides of the equation.

$- 3 x^{2} - 15 x + 42 = 0$

Factor the left side of the equation.

Factor $- 3$ out of $- 3 x^{2} - 15 x + 42$ .

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

$- 3 x^{2} - 15 x + 42 = 0$

Factor $- 3$ out of $- 15 x$ .

$- 3 x^{2} - 3 (5 x) + 42 = 0$

Factor $- 3$ out of $42$ .

$- 3 x^{2} - 3 (5 x) - 3 \cdot - 14 = 0$

Factor $- 3$ out of $- 3 (x^{2}) - 3 (5 x)$ .

$- 3 (x^{2} + 5 x) - 3 \cdot - 14 = 0$

Factor $- 3$ out of $- 3 (x^{2} + 5 x) - 3 (- 14)$ .

$- 3 (x^{2} + 5 x - 14) = 0$

Factor.

Factor $x^{2} + 5 x - 14$ 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 $- 14$ and whose sum is $5$ .

$- 2, 7$

Write the factored form using these integers.

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

Remove unnecessary parentheses.

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

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

$x - 2 = 0$

$x + 7 = 0$

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 $x + 7$ 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 $- 3 (x - 2) (x + 7) = 0$ true.

$x = 2, - 7$

Solve for x (1/8)^(3x^2-3x)=64^(9x-21)

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