Question

1. ** solve: \\(\log_{4}(2x) + \log_{4}(x + 2) = 2\\)

Explanation:

Step1: Apply log addition rule

Using the property $\log_b(M) + \log_b(N) = \log_b(MN)$, we combine the logarithms:
$\log_4(2x(x + 2)) = 2$

Step2: Convert to exponential form

Recall that $\log_b(a) = c$ is equivalent to $b^c = a$. So we have:
$4^2 = 2x(x + 2)$

Step3: Simplify and solve quadratic

Simplify $4^2$ to 16:
$16 = 2x^2 + 4x$
Divide both sides by 2:
$8 = x^2 + 2x$
Rearrange into standard quadratic form:
$x^2 + 2x - 8 = 0$
Factor the quadratic:
$(x + 4)(x - 2) = 0$
Set each factor to zero:
$x + 4 = 0$ or $x - 2 = 0$
So $x = -4$ or $x = 2$

Step4: Check domain

For $\log_4(2x)$ and $\log_4(x + 2)$ to be defined, $2x > 0$ (so $x > 0$) and $x + 2 > 0$ (so $x > -2$). Thus, $x > 0$.
$x = -4$ is invalid (since $-4
ot> 0$), but $x = 2$ is valid (2 > 0 and 2 + 2 = 4 > 0).

Answer:

$x = 2$