Question

1. solve for (x): (log_2(x + 4) + log_2(x + 3) = 1).

\\(x = 2\\)

\\(x = 5\\)

\\(x = -5\\)

\\(x = -2\\)

Explanation:

🆕 New Concept Discovered: Solving Logarithmic Equations
Combine logs and convert to exponential form.

Step 1: Apply the product rule of logarithms

The product rule states that \(\log_b(A) + \log_b(B) = \log_b(A \cdot B)\). We can use this to combine the two logarithmic terms on the left side:

$$ \log_2(x + 4) + \log_2(x + 3) = 1 $$

$$ \log_2((x + 4)(x + 3)) = 1 $$

Step 2: Convert to exponential form

To get rid of the logarithm, we rewrite the equation in its exponential form. Since \(\log_b(Y) = X\) is equivalent to \(b^X = Y\), we have:

$$ (x + 4)(x + 3) = 2^1 $$

$$ (x + 4)(x + 3) = 2 $$

Step 3: Solve the quadratic equation

First, expand the left side:

$$ x^2 + 3x + 4x + 12 = 2 $$

$$ x^2 + 7x + 12 = 2 $$

Subtract \(2\) from both sides to set the quadratic equation to zero:

$$ x^2 + 7x + 10 = 0 $$

Factor the quadratic equation:

$$ (x + 5)(x + 2) = 0 $$

This gives two potential solutions:

$$ x = -5 \quad \text{or} \quad x = -2 $$

Step 4: Check for extraneous solutions

Logarithms are only defined for positive real numbers, so the arguments inside the original logarithms, \((x + 4)\) and \((x + 3)\), must be strictly greater than \(0\).

• For \(x = -5\):

$$ x + 4 = -5 + 4 = -1 \quad (\text{undefined}) $$

Thus, \(x = -5\) is an extraneous solution.

• For \(x = -2\):

$$ x + 4 = -2 + 4 = 2 > 0 \quad (\text{valid}) $$

$$ x + 3 = -2 + 3 = 1 > 0 \quad (\text{valid}) $$

Thus, \(x = -2\) is the only valid solution.

Answer:

\(x = -2\)