Question

which equation is equivalent to \\(\log 4 + \log(x + 2) = 1\\)?\
\\(x - 2 = 10^1\\)\
\\(4x + 8 = 10^1\\)\
\\(4x + 8 = 1\\)\
\\(\frac{4}{x + 2} = 10^1\\)

Explanation:

Step1: Apply log product rule

$\log(4(x+2)) = 1$

Step2: Convert to exponential form

$4(x+2) = 10^1$

Step3: Expand left-hand side

$4x + 8 = 10^1$

Step4: Solve for $x$

$4x = 10 - 8$
$4x = 2$
$x = \frac{2}{4} = \frac{1}{2}$

Answer:

Equivalent equation: $4x + 8 = 10^1$
Solution: $x = \frac{1}{2}$