We are given the equation:

$\log_2(2x^2 - 2x) = 2$

Step 1: Convert the logarithmic equation to an exponential form.

Using the logarithmic identity:

$\log_b(A) = C \quad \text{is equivalent to} \quad A = b^C$

In this case, the base $b = 2$, and the equation becomes:

$2x^2 - 2x = 2^2$

$2x^2 - 2x = 4$

Step 2: Set the equation equal to zero.

Rearrange the equation to form a standard quadratic:

$2x^2 - 2x - 4 = 0$

Step 3: Simplify the quadratic equation.

Divide the entire equation by 2 to simplify:

$x^2 - x - 2 = 0$

Step 4: Factor the quadratic.

Factor the equation:

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

Step 5: Solve for $x$.

Set each factor equal to zero:

$x - 2 = 0 \quad \Rightarrow \quad x = 2$

$x + 1 = 0 \quad \Rightarrow \quad x = -1$

Step 6: Check for extraneous solutions.

Since we are solving a logarithmic equation, we must ensure that the arguments of the logarithm are positive (the domain of the logarithm). The argument of the logarithm is $2x^2 - 2x$. Substituting $x = 2$ and $x = -1$:

• For $x = 2$:

$2(2)^2 - 2(2) = 8 - 4 = 4 \quad (\text{positive})$

• For $x = -1$:

$2(-1)^2 - 2(-1) = 2 + 2 = 4 \quad (\text{positive})$

Both values of $x$ satisfy the domain of the logarithmic function.

Final Answer:

The solutions are:

$x = 2 \quad \text{and} \quad x = -1$

Would you like more details or have any questions about the steps?

Here are 5 related questions for further practice:

1. Solve $\log_2(x^2 - 3x) = 1$.
2. Solve $\log_3(3x^2 - 6x) = 2$.
3. Find the domain of $\log_2(3x^2 - x)$.
4. Solve $\log_4(x^2 - 2x + 1) = 0$.
5. Solve $\log_5(x^2 - 4x + 4) = 1$.

Tip: Always check if the arguments inside logarithmic functions are positive when solving logarithmic equations to avoid extraneous solutions.