Question

a student showed the steps below while solving the equation 14 = \log_{5}(2x - 3) by graphing.\
step 1: \qquad\quad write a system of equations: y_{1} = 14, y_{2} = \log_{5}(2x - 3)\
step 2: \qquad\quad use the change of base formula to rewrite the equations:\
\qquad\qquad\quad y_{1} = \log 14, y_{2} = \frac{\log(2x - 3)}{\log 5}\
step 3: \qquad\quad graph the two equations:

Explanation:

Step1: Analyze the system of equations

We have \( y_1 = 14 \) (a horizontal line) and \( y_2=\log_5(2x - 3) \). To solve the equation \( 14=\log_5(2x - 3) \) graphically, we find the intersection of \( y_1 = 14 \) and \( y_2=\log_5(2x - 3) \).

Step2: Use the change of base formula

The change of base formula for a logarithm \( \log_b a=\frac{\log_c a}{\log_c b} \) (here we use \( c = 10 \) for common logarithm). So \( y_2=\frac{\log(2x - 3)}{\log 5} \) and \( y_1=\log 14\)? Wait, no, \( y_1 = 14 \) is a constant function, not \( \log 14 \). The student made a mistake in Step 2: \( y_1 = 14 \) (a horizontal line with \( y = 14 \)), not \( y_1=\log 14 \).

Step3: Graph analysis

The graph of \( y_1 = 14 \) is a horizontal line parallel to the x - axis at \( y = 14 \). The graph of \( y_2=\log_5(2x - 3) \) is a logarithmic function. For the logarithmic function \( y=\log_5(2x - 3) \), the argument \( 2x-3>0\Rightarrow x > \frac{3}{2}=1.5 \). To find the solution, we need to find the x - value where \( y_1 = 14 \) and \( y_2=\log_5(2x - 3) \) intersect. Let's solve the equation algebraically to check.

Starting from \( 14=\log_5(2x - 3) \), by the definition of logarithm, if \( \log_b a = c \), then \( a=b^c \). So \( 2x-3 = 5^{14} \), then \( 2x=5^{14}+3 \), \( x=\frac{5^{14}+3}{2} \). But from the graph, the student's graph of \( y_1 \) is incorrect because \( y_1 = 14 \) should be a horizontal line at \( y = 14 \), but in the given graph, the horizontal line is at \( y\approx1 \) (from the graph's y - axis markings). So the error is in Step 2 where the student incorrectly rewrote \( y_1 \) as \( \log 14 \) instead of recognizing \( y_1 = 14 \) is a constant function.

Answer:

The error is in Step 2. The function \( y_1 \) should be \( y_1 = 14 \) (a horizontal line with \( y = 14 \)), not \( y_1=\log 14 \). The student misapplied the change of base formula to the constant function \( y_1 = 14 \).