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 $\log_b a = \frac{\log_c a}{\log_c b}$ (where $c>0, c
eq1$). For the left - hand side $\log_4(x + 3)$, using the change of base formula with base 10 (common logarithm), we have $\log_4(x + 3)=\frac{\log(x + 3)}{\log4}$. Let $y_1=\log_4(x + 3)$, so $y_1=\frac{\log(x + 3)}{\log4}$.

Step2: Apply Change of Base to the Right - Hand Side

For the right - hand side $\log_2(2 + x)$, using the change of base formula with base 10, we get $\log_2(2 + x)=\frac{\log(2 + x)}{\log2}$. Let $y_2=\log_2(2 + x)$, so $y_2=\frac{\log(2 + x)}{\log2}$.

Step3: Analyze the Options

• Option 2: $\log x+3$ is incorrect. The argument of the logarithm is $x + 3$, so it should be $\log(x + 3)$, not $\log x+3$. Similarly for the right - hand side.
• Option 3: The expressions for $y_1$ and $y_2$ do not match the change of base application for the given equation.
• Option 4: The expressions $\frac{\log x + 3}{4}$ and $\frac{\log 2+x}{2}$ are incorrect as they do not follow the change of base formula.
• Option 1: $y_1=\frac{\log(x + 3)}{\log4}$ and $y_2=\frac{\log(2 + x)}{\log2}$ is correct as it follows the change of base formula for $\log_4(x + 3)$ and $\log_2(2 + x)$ respectively.

Answer:

A. $y_{1}=\frac{\log(x + 3)}{\log4},y_{2}=\frac{\log(2 + x)}{\log2}$