Question

$log_{2}(x + 6) + log_{2}x = 4$

Explanation:

Step1: Apply log addition rule

Using the property $\log_a M + \log_a N = \log_a(MN)$, we combine the two logarithms:
$\log_2[(x + 6)x] = 4$

Step2: Convert to exponential form

Recall that $\log_a b = c$ implies $a^c = b$. So, for base 2, we have:
$2^4 = x(x + 6)$

Step3: Simplify and solve quadratic

Simplify $2^4 = 16$, so the equation becomes:
$x^2 + 6x - 16 = 0$
Factor the quadratic:
$(x + 8)(x - 2) = 0$
This gives solutions $x = -8$ or $x = 2$.

Step4: Check domain

The argument of a logarithm must be positive. For $\log_2 x$, $x > 0$; for $\log_2(x + 6)$, $x + 6 > 0$ (which is true if $x > 0$). So $x = -8$ is invalid.

Answer:

$x = 2$