Find the values of x for which log2 2x + log2 (x + 4) = 6.

Steps to Solve

1. Combine Logarithmic Expressions

Use the property of logarithms that states $\log_a b + \log_a c = \log_a (bc)$.

So we can rewrite the equation as: $$\log_2 (2x) + \log_2 (x + 4) = \log_2 (2x(x + 4)).$$

The equation becomes: $$\log_2 (2x(x + 4)) = 6.$$

2. Convert Logarithm to Exponential Form

To eliminate the logarithm, we convert the logarithmic equation to its exponential form: $$2^6 = 2x(x + 4).$$

Calculating $2^6$ gives us: $$64 = 2x(x + 4).$$

3. Expand and Rearrange the Equation

Expanding the right side, we have: $$64 = 2x^2 + 8x.$$

To rearrange, set the equation to 0: $$2x^2 + 8x - 64 = 0.$$

4. Divide through by 2

To simplify, divide the entire equation by 2: $$x^2 + 4x - 32 = 0.$$

5. Factor the Quadratic Equation

Next, we will factor the quadratic expression: $$(x + 8)(x - 4) = 0.$$

6. Solve for x

Setting each factor to zero gives us the solutions: $$x + 8 = 0 \quad \Rightarrow \quad x = -8,$$ $$x - 4 = 0 \quad \Rightarrow \quad x = 4.$$

7. Check for Valid Solutions

Since logarithms are only defined for positive arguments, we check:

• For $x = -8$: $\log_2(2 \cdot -8)$ and $\log_2(-8 + 4)$ are not defined.
• For $x = 4$: Both $\log_2(2 \cdot 4)$ and $\log_2(4 + 4)$ are valid.

Thus, the only valid solution is $x = 4$.