A student showed the steps below while solving the equation $14=\log _5(2 x-3)$ by graphing.
Step 1:
Step 2:
Step 3:
Write a system of equations: $y_1=14, y_2=\log _5(2 x-3)$
Use the change of base formula to rewrite the equations:
$y_1=\log 14 ; y_2=\frac{\log (2 x-3)}{\log 5}$
Graph the two equations:
Answer (2)
The student incorrectly applies the change of base formula to the constant term.
The correct system of equations should be y 1 = 14 and y 2 = l o g 5 l o g ( 2 x − 3 ) .
The equation 14 = lo g 5 ( 2 x − 3 ) can be rewritten as 5 14 = 2 x − 3 .
Solving for x , we get x = 2 5 14 + 3 = 3051757814 , so the solution is 3051757814 .
Explanation
Analyzing the Problem The student is trying to solve the equation 14 = lo g 5 ( 2 x − 3 ) by graphing. The student sets up a system of equations y 1 = 14 and y 2 = lo g 5 ( 2 x − 3 ) . The student then attempts to use the change of base formula but makes an error.
Identifying the Error The change of base formula is correctly applied to y 2 = lo g 5 ( 2 x − 3 ) to get y 2 = l o g 5 l o g ( 2 x − 3 ) . However, the student incorrectly applies the change of base formula to y 1 = 14 . The number 14 is a constant and does not need the change of base formula.
Correcting the Equations The correct system of equations to graph is y 1 = 14 and y 2 = l o g 5 l o g ( 2 x − 3 ) . The student incorrectly rewrites y 1 = 14 as y 1 = lo g 14 . This is where the error occurs.
Solving Algebraically To solve the equation algebraically, we can rewrite the original equation 14 = lo g 5 ( 2 x − 3 ) as 5 14 = 2 x − 3 . Then, we solve for x :
2 x = 5 14 + 3
x = 2 5 14 + 3
Calculating this value, we have:
5 14 = 6103515625
x = 2 6103515625 + 3 = 2 6103515628 = 3051757814
So, x = 3051757814 .
Conclusion The student's mistake is in Step 2, where they change y 1 = 14 to y 1 = lo g 14 . This is incorrect. The correct approach is to graph y 1 = 14 and y 2 = l o g 5 l o g ( 2 x − 3 ) and find their intersection. The x-coordinate of the intersection point will be the solution to the original equation.
Examples
Logarithmic equations are used in various fields such as calculating the magnitude of earthquakes on the Richter scale, measuring the intensity of sound (decibels), and determining the pH of a solution in chemistry. Understanding how to solve logarithmic equations is crucial for making accurate measurements and predictions in these areas. For example, if we know the intensity of an earthquake is I and the intensity of a reference earthquake is I 0 , the magnitude M on the Richter scale is given by M = lo g 10 ( I 0 I ) . Solving logarithmic equations allows us to determine the intensity of an earthquake given its magnitude.
The student incorrectly applied the change of base formula to the constant term while solving the equation 14 = lo g 5 ( 2 x − 3 ) . The correct approach is to graph y 1 = 14 and y 2 = l o g 5 l o g ( 2 x − 3 ) , leading to the solution x = 3051757814 .
;
