Let's solve each equation step by step.

(a) $\ln(x) + \ln(x - 3) = 0$

First, we use the logarithmic property that states:

$\ln(a) + \ln(b) = \ln(a \times b)$

So, the equation becomes:

$\ln(x \times (x - 3)) = 0$

Now, to solve for $x$, we exponentiate both sides to get rid of the logarithm:

$x(x - 3) = e^0$

Since $e^0 = 1$, the equation simplifies to:

$x(x - 3) = 1$

This expands to:

$x^2 - 3x - 1 = 0$

This is a quadratic equation. We solve it using the quadratic formula:

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

Here, $a = 1$, $b = -3$, and $c = -1$. Plugging in these values:

$x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(1)(-1)}}{2(1)}$

$x = \frac{3 \pm \sqrt{9 + 4}}{2}$

$x = \frac{3 \pm \sqrt{13}}{2}$

So, the exact solutions are:

$x = \frac{3 + \sqrt{13}}{2} \quad \text{or} \quad x = \frac{3 - \sqrt{13}}{2}$

Since $x$ must be greater than 3 for $\ln(x - 3)$ to be defined, we discard the solution $\frac{3 - \sqrt{13}}{2}$.

Exact value:
$x = \frac{3 + \sqrt{13}}{2}$

Decimal approximation:
Calculate $\frac{3 + \sqrt{13}}{2}$:

$x \approx \frac{3 + 3.606}{2} \approx \frac{6.606}{2} \approx 3.303$

(b) $51 - 6x = 9$

Let's solve this linear equation for $x$.

Subtract 51 from both sides:

$-6x = 9 - 51$

$-6x = -42$

Now, divide both sides by $-6$:

$x = \frac{-42}{-6} = 7$

Exact value:
$x = 7$

Decimal approximation:
$x = 7.000$ (as the exact value is an integer, the decimal approximation is the same).

Final Answers

(a) $\ln(x) + \ln(x - 3) = 0$
Exact value: $x = \frac{3 + \sqrt{13}}{2}$
Decimal approximation: $x \approx 3.303$

(b) $51 - 6x = 9$
Exact value: $x = 7$
Decimal approximation: $x = 7.000$

Would you like more details or have any questions about these solutions?

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Here are 5 related questions:

1. How do you derive the quadratic formula?
2. Can the quadratic formula give complex solutions, and when?
3. How do you interpret the natural logarithm function graphically?
4. What is the significance of the discriminant in a quadratic equation?
5. How can you check if your solutions to a logarithmic equation are valid?

Tip: When solving logarithmic equations, always consider the domain restrictions, as logarithms are only defined for positive arguments.