Solve the Logarithmic Equation: log3x + log(x-1) = 3

To solve the problem, we'll follow these steps:

• Step 1: Combine the logarithms using the product rule.
• Step 2: Convert the logarithmic equation to an exponential equation.
• Step 3: Simplify and solve the quadratic equation.
• Step 4: Consider only solutions greater than 1.

Now, let's work through each step:
Step 1: Combine the logarithms using the product rule:
$\log(3x) + \log(x-1) = \log((3x)(x-1))$.
Step 2: Convert the logarithmic equation to an exponential equation:
$\log((3x)(x-1)) = 3 \Rightarrow (3x)(x-1) = 10^3$.
Step 3: Simplify the quadratic equation:
$(3x)(x-1) = 1000$:
$3x^2 - 3x = 1000$.
$\Rightarrow 3x^2 - 3x - 1000 = 0$.
Divide by 3 to simplify:
$x^2 - x - \frac{1000}{3} = 0$.
Solve this equation using the quadratic formula:
The quadratic formula is $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.
Here, $a = 1$, $b = -1$, and $c = -\frac{1000}{3}$.
Calculate the discriminant:\ $D = (-1)^2 - 4 \times 1 \times \left(-\frac{1000}{3}\right)\ = 1 + \frac{4000}{3} = \frac{4003}{3}$.
Now, calculate $x$:
$x = \frac{-(-1) \pm \sqrt{\frac{4003}{3}}}{2 \times 1}$.
$\Rightarrow x = \frac{1 \pm \sqrt{\frac{4003}{3}}}{2}$.
Calculating this gives approximately $x \approx 18.8$.
Step 4: Verify that $x > 1$ to be in the domain.
Since this is true, the valid solution is within the domain, confirming:
Therefore, the solution to the problem is $x = 18.8$.