a) Prove that the following expression is a rational number: $\sqrt{3+2\sqrt{2}} - \sqrt{3-2\sqrt{2}}$ b) Find the solution set for the equation $2 \log x = \log 4 + \log (2x-3)$

Steps to Solve

1. Part (a): Square the expression

Let $A = \sqrt{3+2\sqrt{2}} - \sqrt{3-2\sqrt{2}}$. Squaring both sides, we get: $A^2 = (\sqrt{3+2\sqrt{2}} - \sqrt{3-2\sqrt{2}})^2$ $A^2 = (3+2\sqrt{2}) - 2 \sqrt{(3+2\sqrt{2})(3-2\sqrt{2})} + (3-2\sqrt{2})$

2. Simplify the squared expression

Simplifying the above equation: $A^2 = 3 + 2\sqrt{2} - 2\sqrt{9 - (4 \times 2)} + 3 - 2\sqrt{2}$ $A^2 = 6 - 2\sqrt{9-8}$ $A^2 = 6 - 2\sqrt{1}$ $A^2 = 6 - 2(1)$ $A^2 = 6 - 2$ $A^2 = 4$

3. Solve for A

Taking the square root of both sides: $A = \pm \sqrt{4}$ $A = \pm 2$ Since $\sqrt{3+2\sqrt{2}} > \sqrt{3-2\sqrt{2}}$, $A$ must be positive. Therefore, $A = 2$.

4. Conclusion for part (a) Since $A = 2$, which is a rational number, the given expression is a rational number.

5. Part (b): Use logarithm properties to simplify the equation

The given equation is $2 \log x = \log 4 + \log (2x-3)$.

Using the logarithm property $n \log a = \log a^n$, we have: $\log x^2 = \log 4 + \log (2x-3)$

Using the logarithm property $\log a + \log b = \log (ab)$, we have: $\log x^2 = \log [4(2x-3)]$

6. Remove the logarithms

Since the logarithms are equal, their arguments must be equal: $x^2 = 4(2x-3)$ $x^2 = 8x - 12$

7. Solve the quadratic equation

Rearrange the equation into a quadratic equation: $x^2 - 8x + 12 = 0$

Factor the quadratic equation: $(x-6)(x-2) = 0$

So, $x = 6$ or $x = 2$.

8. Check for extraneous solutions

We need to make sure that the solutions are valid in the original logarithmic equation $2 \log x = \log 4 + \log (2x-3)$. The argument of any logarithm must be positive. Thus $x>0$ and $2x-3>0$ which implies $x>\frac{3}{2}$.

For $x=6$: $2 \log 6 = \log 4 + \log (2(6)-3)$ $2 \log 6 = \log 4 + \log 9$ $\log 36 = \log (4 \times 9)$ $\log 36 = \log 36$. This is true.

For $x=2$: $2 \log 2 = \log 4 + \log (2(2)-3)$ $2 \log 2 = \log 4 + \log 1$ $\log 4 = \log 4 + 0$ $\log 4 = \log 4$. This is true. Since both solutions satisfy $x > \frac{3}{2}$, both are valid.