Solve for x log base 4 of 2x^2-4x=2

$\log_{4} (2 x^{2} - 4 x) = 2$

Rewrite $\log_{4} (2 x^{2} - 4 x) = 2$ in exponential form using the definition of a logarithm. If $x$ and $b$ are positive real numbers and $b \neq 1$ , then $\log_{b} (x) = y$ is equivalent to $b^{y} = x$ .

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

Solve for $x$ .

Rewrite the equation as $2 x^{2} - 4 x = 4^{2}$ .

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

Raise $4$ to the power of $2$ .

$2 x^{2} - 4 x = 16$

Subtract $16$ from both sides of the equation.

$2 x^{2} - 4 x - 16 = 0$

Factor the left side of the equation.

Factor $2$ out of $2 x^{2} - 4 x - 16$ .

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

$2 (x^{2}) - 4 x - 16 = 0$

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

$2 (x^{2}) + 2 (- 2 x) - 16 = 0$

Factor $2$ out of $- 16$ .

$2 x^{2} + 2 (- 2 x) + 2 \cdot - 8 = 0$

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

$2 (x^{2} - 2 x) + 2 \cdot - 8 = 0$

Factor $2$ out of $2 (x^{2} - 2 x) + 2 \cdot - 8$ .

$2 (x^{2} - 2 x - 8) = 0$

Factor.

Factor $x^{2} - 2 x - 8$ 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 $- 8$ and whose sum is $- 2$ .

$- 4, 2$

Write the factored form using these integers.

$2 ((x - 4) (x + 2)) = 0$

Remove unnecessary parentheses.

$2 (x - 4) (x + 2) = 0$

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

$x - 4 = 0$

$x + 2 = 0$

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 $x + 2$ equal to $0$ and solve for $x$ .

Set $x + 2$ equal to $0$ .

$x + 2 = 0$

Subtract $2$ from both sides of the equation.

$x = - 2$

The final solution is all the values that make $2 (x - 4) (x + 2) = 0$ true.

$x = 4, - 2$

Solve for x log base 4 of 2x^2-4x=2

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