Solve for x (1/3)^(2x^2-10x)=(1/27)^(-2x+42)

${(\frac{1}{3})}^{2 x^{2} - 10 x} = {(\frac{1}{27})}^{- 2 x + 42}$

Step 1

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

${(\frac{1}{3})}^{2 x^{2} - 10 x} = {(\frac{1}{3})}^{3 (- 2 x + 42)}$

Step 2

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

$2 x^{2} - 10 x = 3 (- 2 x + 42)$

Step 3

Solve for $x$ .

Simplify $3 (- 2 x + 42)$ .

Apply the distributive property.

$2 x^{2} - 10 x = 3 (- 2 x) + 3 \cdot 42$

Multiply.

Multiply $- 2$ by $3$ .

$2 x^{2} - 10 x = - 6 x + 3 \cdot 42$

Multiply $3$ by $42$ .

$2 x^{2} - 10 x = - 6 x + 126$

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

Add $6 x$ to both sides of the equation.

$2 x^{2} - 10 x + 6 x = 126$

Add $- 10 x$ and $6 x$ .

$2 x^{2} - 4 x = 126$

Subtract $126$ from both sides of the equation.

$2 x^{2} - 4 x - 126 = 0$

Factor the left side of the equation.

Factor $2$ out of $2 x^{2} - 4 x - 126$ .

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

$2 (x^{2}) - 4 x - 126 = 0$

Factor $2$ out of $- 4 x$ .

$2 (x^{2}) + 2 (- 2 x) - 126 = 0$

Factor $2$ out of $- 126$ .

$2 x^{2} + 2 (- 2 x) + 2 \cdot - 63 = 0$

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

$2 (x^{2} - 2 x) + 2 \cdot - 63 = 0$

Factor $2$ out of $2 (x^{2} - 2 x) + 2 \cdot - 63$ .

$2 (x^{2} - 2 x - 63) = 0$

Factor.

Factor $x^{2} - 2 x - 63$ 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 $- 63$ and whose sum is $- 2$ .

$- 9, 7$

Write the factored form using these integers.

$2 ((x - 9) (x + 7)) = 0$

Remove unnecessary parentheses.

$2 (x - 9) (x + 7) = 0$

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

$x - 9 = 0$

$x + 7 = 0$

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

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

$x - 9 = 0$

Add $9$ to both sides of the equation.

$x = 9$

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 $2 (x - 9) (x + 7) = 0$ true.

$x = 9, - 7$

Solve for x (1/3)^(2x^2-10x)=(1/27)^(-2x+42)

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