If A = {x ∈ R : 4x² = 2x + 3} and B = {x ∈ R : log(x²) = (log(x))²}, find A ∩ B.

Steps to Solve

1. Solve for set A

To solve for set A, we start with the equation:

$$ 4x^2 = 2x + 3 $$

Rearranging all terms to one side gives:

$$ 4x^2 - 2x - 3 = 0 $$

Next, we can use the quadratic formula, which is:

$$ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$

Here, (a = 4), (b = -2), and (c = -3).

Calculating the discriminant:

$$ b^2 - 4ac = (-2)^2 - 4 \cdot 4 \cdot (-3) = 4 + 48 = 52 $$

Now we can find the roots:

$$ x = \frac{-(-2) \pm \sqrt{52}}{2 \cdot 4} = \frac{2 \pm 2\sqrt{13}}{8} = \frac{1 \pm \sqrt{13}}{4} $$

The solutions for set A are:

$$ x = \frac{1 + \sqrt{13}}{4}, \quad x = \frac{1 - \sqrt{13}}{4} $$

2. Solve for set B

For set B, we have the equation:

$$ \log(x^2) = (\log(x))^2 $$

Using properties of logarithms, we can rewrite the left side:

$$ \log(x^2) = 2\log(x) $$

Setting the equations equal, we have:

$$ 2\log(x) = (\log(x))^2 $$

Rearranging gives:

$$ (\log(x))^2 - 2\log(x) = 0 $$

Factoring this equation yields:

$$ \log(x)(\log(x) - 2) = 0 $$

Thus, we have:

1. ( \log(x) = 0 ) which gives ( x = 10^0 = 1 )
2. ( \log(x) - 2 = 0 ) which gives ( x = 10^2 = 100 )

Hence, the solutions for set B are:

$$ x = 1, \quad x = 100 $$

3. Find the intersection ( A \cap B )

Now, we need to check which solutions from set A are also in set B:

• For ( x = \frac{1 + \sqrt{13}}{4} ): Calculate approximately:

$$ \sqrt{13} \approx 3.6 \implies \frac{4.6}{4} \approx 1.15 $$

This is not equal to 1 or 100.

• For ( x = \frac{1 - \sqrt{13}}{4} ):

This value is negative and thus not valid in set B.

Therefore, both values from set A are not in set B, and the intersection is empty.