Which of the following shows the extraneous solution to the logarithmic equation [tex]$\log _7\left(3 x^3+x\right)-\log _7(x)=2$[/tex]?
A. [tex]$x=-16$[/tex]
B. [tex]$x=-4$[/tex]
C. [tex]$x=4$[/tex]
D. [tex]$x=16$[/tex]

Use the logarithm property to rewrite the equation as lo g 7 ​ ( x 3 x 3 + x ​ ) = 2 .
Simplify the equation to lo g 7 ​ ( 3 x 2 + 1 ) = 2 .
Convert to exponential form: 3 x 2 + 1 = 7 2 , which simplifies to x 2 = 16 .
Solve for x to get x = ± 4 . Since the original equation contains lo g 7 ​ ( x ) , x must be positive, so the extraneous solution is x = − 4 ​ .

Explanation

Understanding the Problem We are given the logarithmic equation lo g 7 ​ ( 3 x 3 + x ) − lo g 7 ​ ( x ) = 2 . We need to find the extraneous solution to this equation from the given options: x = − 16 , x = − 4 , x = 4 , x = 16 . An extraneous solution is a solution that arises from the process of solving the equation but is not a valid solution to the original equation. In logarithmic equations, extraneous solutions often occur when a potential solution results in taking the logarithm of a non-positive number.

Applying Logarithmic Properties First, we use the logarithm property lo g a ​ ( b ) − lo g a ​ ( c ) = lo g a ​ ( c b ​ ) to rewrite the equation as lo g 7 ​ ( x 3 x 3 + x ​ ) = 2 .

Simplifying the Equation Next, we simplify the argument of the logarithm: lo g 7 ​ ( 3 x 2 + 1 ) = 2 . This simplification is valid only if x  = 0 .

Converting to Exponential Form Now, we convert the logarithmic equation to an exponential equation: 3 x 2 + 1 = 7 2 .

Solving for x We solve for x : 3 x 2 + 1 = 49 , so 3 x 2 = 48 , and x 2 = 16 . Thus, x = ± 4 .

Identifying Extraneous Solutions We need to check the solutions in the original equation. Since we have lo g 7 ​ ( x ) in the original equation, x must be positive. Therefore, x = − 4 is an extraneous solution because we cannot take the logarithm of a negative number.

Verifying the Valid Solution We verify that x = 4 is a valid solution. lo g 7 ​ ( 3 ( 4 ) 3 + 4 ) − lo g 7 ​ ( 4 ) = lo g 7 ​ ( 192 + 4 ) − lo g 7 ​ ( 4 ) = lo g 7 ​ ( 196 ) − lo g 7 ​ ( 4 ) = lo g 7 ​ ( 4 196 ​ ) = lo g 7 ​ ( 49 ) = 2 . So x = 4 is a valid solution.

Conclusion Therefore, the extraneous solution is x = − 4 .


Examples
Logarithmic equations are used in various fields such as calculating the magnitude of earthquakes on the Richter scale, determining the acidity or alkalinity (pH) of a solution in chemistry, and modeling population growth or decay in biology. Understanding how to solve logarithmic equations and identify extraneous solutions is crucial in these applications to ensure accurate and meaningful results. For instance, in seismology, an extraneous solution could lead to an incorrect estimation of an earthquake's strength, which could have serious implications for disaster response and preparedness.

Answered by GinnyAnswer | 2025-07-05

The extraneous solution to the logarithmic equation lo g 7 ​ ( 3 x 3 + x ) − lo g 7 ​ ( x ) = 2 is x = − 4 since logarithmic functions cannot take negative arguments. The other potential solution, x = 4 , is valid. Therefore, from the options given, the correct choice is x = − 4 .
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Answered by Anonymous | 2025-07-07