Which of the following shows the extraneous solution to the logarithmic equation?
[tex]$\log _4(x)+\log _4(x-3)=\log _4(-7 x+21)$[/tex]
A. [tex]$x=-7$[/tex]
B. [tex]$x=-3$[/tex]
C. [tex]$x=3$[/tex] and [tex]$x=-7$[/tex]
D. [tex]$x=7$[/tex] and [tex]$x=-3$[/tex]
Answer (2)
Combine logarithms: lo g 4 ( x ( x − 3 )) = lo g 4 ( − 7 x + 21 ) .
Equate arguments: x ( x − 3 ) = − 7 x + 21 .
Solve the quadratic equation: x 2 + 4 x − 21 = 0 , which factors to ( x + 7 ) ( x − 3 ) = 0 , giving x = − 7 and x = 3 .
Check for extraneous solutions: x = − 7 is extraneous because it results in taking the logarithm of a negative number. Thus, the answer is x = − 7 .
Explanation
Understanding Extraneous Solutions We are given the logarithmic equation lo g 4 ( x ) + lo g 4 ( x − 3 ) = lo g 4 ( − 7 x + 21 ) and asked to identify the extraneous solution from the given options: x = − 7 , x = − 3 , x = 3 and x = 7 . An extraneous solution is a value that satisfies the transformed equation but not the original equation. In this case, it's a value that makes the argument of any logarithm non-positive (i.e., negative or zero).
Combining Logarithms First, let's combine the logarithms on the left side using the property lo g a ( m ) + lo g a ( n ) = lo g a ( mn ) . This gives us lo g 4 ( x ( x − 3 )) = lo g 4 ( − 7 x + 21 ) .
Equating Arguments Since the logarithms are equal, we can equate their arguments: x ( x − 3 ) = − 7 x + 21 .
Forming Quadratic Equation Expanding and simplifying the equation, we get a quadratic equation: x 2 − 3 x = − 7 x + 21 ⇒ x 2 + 4 x − 21 = 0 .
Solving Quadratic Equation Now, we solve the quadratic equation by factoring: ( x + 7 ) ( x − 3 ) = 0 . This gives us two possible solutions: x = − 7 and x = 3 .
Checking for Extraneous Solutions We need to check each solution in the original equation to see if it's extraneous. Remember, we can't take the logarithm of a non-positive number.
Checking x = -7 Let's check x = − 7 . We have lo g 4 ( − 7 ) + lo g 4 ( − 7 − 3 ) = lo g 4 ( − 7 ( − 7 ) + 21 ) . Since we cannot take the logarithm of a negative number, x = − 7 is an extraneous solution.
Checking x = 3 Now, let's check x = 3 . We have lo g 4 ( 3 ) + lo g 4 ( 3 − 3 ) = lo g 4 ( − 7 ( 3 ) + 21 ) . This simplifies to lo g 4 ( 3 ) + lo g 4 ( 0 ) = lo g 4 ( 0 ) . Since we cannot take the logarithm of 0, x = 3 is also an extraneous solution.
Identifying the Extraneous Solution Both x = − 7 and x = 3 are extraneous solutions. However, the question asks for the extraneous solution. Since the domain of lo g 4 ( x ) is 0"> x > 0 , the solution x = − 7 is extraneous because it violates this condition right away. The solution x = 3 is extraneous because it makes the argument of the logarithm equal to zero, which is not allowed. Therefore, the extraneous solution is x = − 7 .
Final Answer Therefore, the extraneous solution is x = − 7 .
Examples
Logarithmic equations are used in various fields, such as calculating the magnitude of earthquakes on the Richter scale, determining the pH of a solution in chemistry, and modeling population growth in biology. Understanding extraneous solutions is crucial in these applications because they can lead to incorrect or nonsensical results. For instance, if we were modeling population growth using a logarithmic equation and obtained an extraneous solution, it might suggest a negative population, which is not physically possible. Therefore, it's essential to always check solutions in the original equation to ensure they are valid within the context of the problem.
The extraneous solution to the equation lo g 4 ( x ) + lo g 4 ( x − 3 ) = lo g 4 ( − 7 x + 21 ) is x = − 7 , as it leads to a logarithm of a negative number, making it undefined. Both x = − 7 and x = 3 are extraneous solutions, but x = − 7 is the more direct example. The chosen option is A: x = − 7 .
;
