Question

consider the equation below.
$log_{4}(x+3) = log_{2}(2+x)$
which system of equations can represent the equation?
$\bigcirc y_1 = \frac{\log(x+3)}{\log4}, y_2 = \frac{\log(2+x)}{\log2}$
$\bigcirc y_1 = \frac{\log x+3}{\log4}, y_2 = \frac{\log2+x}{\log2}$
$\bigcirc y_1 = \frac{\log4}{\log2}, y_2 = \frac{\log(x+3)}{\log(2+x)}$
$\bigcirc y_1 = \frac{\log x+3}{4}, y_2 = \frac{\log2+x}{2}$

Explanation:

Step1: Recall change of base formula

The change of base formula for logarithms is $\log_b(a) = \frac{\log(a)}{\log(b)}$, where the new base is a positive number not equal to 1 (here we use common log, base 10).

Step2: Apply formula to left side

For $\log_4(x+3)$, substitute $b=4$, $a=x+3$:
$\log_4(x+3) = \frac{\log(x+3)}{\log4}$

Step3: Apply formula to right side

For $\log_2(2+x)$, substitute $b=2$, $a=2+x$:
$\log_2(2+x) = \frac{\log(2+x)}{\log2}$

Step4: Set as system of equations

To represent the original equation, set $y_1$ equal to the left-hand side and $y_2$ equal to the right-hand side.

Answer:

$y_1 = \frac{\log(x+3)}{\log4}, y_2 = \frac{\log(2+x)}{\log2}$ (the first option)