Step-by-step solution of a logarithmic equation leading to quadratic form and final solutions.

Let’s solve the equation step by step:

We are given:
> log₂(x) + log₂(x - 3) = 2

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Step 1: Use logarithm rule

When you add two logs with the same base, you can multiply their arguments:

> log₂(x) + log₂(x - 3) = log₂[ x · (x - 3) ]

So the equation becomes:

> log₂(x² - 3x) = 2

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Step 2: Convert from log form to exponential form

Recall:
If log_b(A) = C, then A = b^C

So here:

> x² - 3x = 2²
> x² - 3x = 4

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Step 3: Move all terms to one side to make a quadratic equation

Subtract 4 from both sides:

> x² - 3x - 4 = 0

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Step 4: Factor the quadratic

We need two numbers that multiply to -4 and add to -3.

Those numbers are -4 and +1.

So:

> (x - 4)(x + 1) = 0

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Step 5: Solve for x

Set each factor equal to zero:

- x - 4 = 0 → x = 4
- x + 1 = 0 → x = -1

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Step 6: Check for extraneous solutions

Logarithms are only defined for positive numbers. So we must check if these values make the original logs valid.

Original equation: log₂(x) + log₂(x - 3) = 2

Check x = 4:

- log₂(4) → OK (4 > 0)
- log₂(4 - 3) = log₂(1) → OK (1 > 0)

✔ Valid solution.

Check x = -1:

- log₂(-1) → ✘ Not allowed! Log of negative number is undefined.
- Also, log₂(-1 - 3) = log₂(-4) → also invalid.

✘ So x = -1 is NOT a valid solution.

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Only x = 4 works.

Final Answer:
4