Solve for x log base 2x of 2x^2+6x-4=2

$\log_{2 x} (2 x^{2} + 6 x - 4) = 2$

Rewrite $\log_{2 x} (2 x^{2} + 6 x - 4) = 2$ 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 $b^{y} = x$ .

${(2 x)}^{2} = 2 x^{2} + 6 x - 4$

Solve for $x$ .

Since $x$ is on the right side of the equation, switch the sides so it is on the left side of the equation.

$2 x^{2} + 6 x - 4 = {(2 x)}^{2}$

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

Apply the product rule to $2 x$ .

$2 x^{2} + 6 x - 4 = 2^{2} x^{2}$

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

$2 x^{2} + 6 x - 4 = 4 x^{2}$

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

Subtract $4 x^{2}$ from both sides of the equation.

$2 x^{2} + 6 x - 4 - 4 x^{2} = 0$

Subtract $4 x^{2}$ from $2 x^{2}$ .

$- 2 x^{2} + 6 x - 4 = 0$

Factor the left side of the equation.

Factor $- 2$ out of $- 2 x^{2} + 6 x - 4$ .

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

$- 2 x^{2} + 6 x - 4 = 0$

Factor $- 2$ out of $6 x$ .

$- 2 x^{2} - 2 (- 3 x) - 4 = 0$

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

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

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

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

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

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

Factor.

Factor $x^{2} - 3 x + 2$ 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 $2$ and whose sum is $- 3$ .

$- 2, - 1$

Write the factored form using these integers.

$- 2 ((x - 2) (x - 1)) = 0$

Remove unnecessary parentheses.

$- 2 (x - 2) (x - 1) = 0$

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

$x - 2 = 0$

$x - 1 = 0$

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 $x - 1$ equal to $0$ and solve for $x$ .

Set $x - 1$ equal to $0$ .

$x - 1 = 0$

Add $1$ to both sides of the equation.

$x = 1$

The final solution is all the values that make $- 2 (x - 2) (x - 1) = 0$ true.

$x = 2, 1$

Solve for x log base 2x of 2x^2+6x-4=2

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