Which of the following shows the extraneous solution to the logarithmic equation?
$\log _4(x)+\log _4(x-3)=\log _4(-7 x+21)$
A. $x=-7$
B. $x=-3$
C. $x=3$ and $x=-7$
D. $x=7$ and $x=-3$
Answer (1)
Combine logarithms using the property lo g a ( b ) + lo g a ( c ) = lo g a ( b c ) .
Equate the arguments of the logarithms.
Solve the resulting quadratic equation.
Check the solutions in the original equation to identify extraneous solutions. The extraneous solution is x = − 7 .
Explanation
Understanding the Problem We are given the logarithmic equation lo g 4 ( x ) + lo g 4 ( x − 3 ) = lo g 4 ( − 7 x + 21 ) and asked to find the extraneous solution. Extraneous solutions are solutions that we find algebraically but do not satisfy the original equation. In the context of logarithmic equations, this usually means that the solution makes the argument of a logarithm negative or zero.
Combining Logarithms First, we combine the logarithms on the left side using the property lo g a ( b ) + lo g a ( c ) = lo g a ( b c ) . This gives us lo g 4 ( x ( x − 3 )) = lo g 4 ( − 7 x + 21 ) .
Equating Arguments Since the logarithms have the same base, we can equate the arguments: x ( x − 3 ) = − 7 x + 21 .
Simplifying to Quadratic Form Expanding and simplifying the equation, we get a quadratic equation: x 2 − 3 x = − 7 x + 21 , which simplifies to x 2 + 4 x − 21 = 0 .
Solving the Quadratic Equation We can factor the quadratic equation as ( x + 7 ) ( x − 3 ) = 0 . Thus, the possible solutions are x = − 7 and x = 3 .
Checking for Extraneous Solutions Now we need to check each solution in the original equation to see if it is extraneous. Remember, the argument of a logarithm must be positive.
Checking x = -7 Let's check x = − 7 . We have lo g 4 ( − 7 ) , which is undefined since the argument is negative. Therefore, 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 ( 3 ) + lo g 4 ( 0 ) . Since lo g 4 ( 0 ) is undefined, x = 3 is also an extraneous solution.
Identifying the Extraneous Solution Since both x = − 7 and x = 3 are extraneous solutions, and the problem asks for the extraneous solution, we need to determine which one is considered the extraneous solution. In this case, x = − 7 makes the argument of the first logarithm negative, which is a direct violation of the logarithm's domain. Therefore, x = − 7 is the extraneous solution.
Final Answer The extraneous solution is x = − 7 .
Examples
Logarithmic equations are used in various fields such as physics, engineering, and finance. For example, in radioactive decay, the amount of a radioactive substance remaining after a certain time is modeled using a logarithmic equation. Similarly, in finance, the growth of an investment can be modeled using exponential and logarithmic functions. Understanding how to solve logarithmic equations and identify extraneous solutions is crucial in these applications to ensure accurate and meaningful results.
