Question

which system of equations could be graphed to solve the equation below?
\\(\log_{0.5} x = \log_{3} 2 + x\\)
\\(\bigcirc\\) \\(y_1 = \frac{\log 0.5}{x}, y_2 = \frac{\log 3}{2 + x}\\)
\\(\bigcirc\\) \\(y_1 = \frac{\log x}{\log 0.5}, y_2 = \frac{\log 2 + x}{\log 3}\\)
\\(\bigcirc\\) \\(y_1 = \frac{\log 0.5}{\log 3}, y_2 = \frac{\log x}{\log 2 + x}\\)
\\(\bigcirc\\) \\(y_1 = \frac{\log x}{\log 0.5}, y_2 = \frac{\log 2}{\log 3} + x\\)

Explanation:

Step1: Recall Change of Base Formula

The change of base formula for logarithms is $\log_{a}b=\frac{\log b}{\log a}$ (where the base of the new logarithm can be any positive number not equal to 1, often we use base 10 or base $e$).

For the left - hand side of the equation $\log_{0.5}x$, using the change of base formula, we have $\log_{0.5}x = \frac{\log x}{\log 0.5}$. So we can let $y_1=\frac{\log x}{\log 0.5}$.

Step2: Simplify the Right - hand Side

For the right - hand side of the equation $\log_{3}2 + x$, using the change of base formula on $\log_{3}2$, we get $\log_{3}2=\frac{\log 2}{\log 3}$. So the right - hand side is $\frac{\log 2}{\log 3}+x$. We can let $y_2 = \frac{\log 2}{\log 3}+x$.

Answer:

$y_1=\frac{\log x}{\log 0.5},y_2=\frac{\log 2}{\log 3}+x$ (the last option among the given choices)