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]
Answer (1)
∙ Use the logarithm property to rewrite the equation as lo g 7 ( x 3 x 3 + x ) = 2 .$\bullet S im pl i f y t h ee q u a t i o n t o \log_7(3x^2+1) = 2 .$ ∙ Convert to exponential form: 3 x 2 + 1 = 49 , and solve for x , obtaining x = ± 4 .$\bullet C h ec k t h eso l u t i o n s in t h eor i g ina l e q u a t i o n . S in ce \log_7(x) m u s t ha v e a p os i t i v e a r gu m e n t , x = -4 i s an e x t r an eo u sso l u t i o n . T h ee x t r an eo u sso l u t i o ni s \boxed{x=-4}$.
Explanation
Understanding the Problem We are given the logarithmic equation lo g 7 ( 3 x 3 + x ) − lo g 7 ( x ) = 2 and asked to find the extraneous solution from the options x = − 16 , x = − 4 , x = 4 , x = 16 . An extraneous solution is a solution that arises during the solving process but does not satisfy the original equation. Logarithmic functions are only defined for positive arguments. Therefore, we must check if the given solutions satisfy the condition that the arguments of the logarithms are positive.
Applying Logarithm 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 .
Converting to Exponential Form Now, we convert the logarithmic equation to an exponential equation: 3 x 2 + 1 = 7 2 , which simplifies to 3 x 2 + 1 = 49 .
Solving for x Solving for x : 3 x 2 = 48 , so x 2 = 16 , which gives 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 the logarithm of a negative number is undefined.
Verifying Valid Solutions We verify that x = 4 is a valid solution. If x = 4 , then 0"> 3 x 3 + x = 3 ( 4 3 ) + 4 = 3 ( 64 ) + 4 = 192 + 4 = 196 > 0 . So lo g 7 ( 3 x 3 + x ) is defined. Also, lo g 7 ( x ) = lo g 7 ( 4 ) is defined. Thus, x = 4 is a valid solution.
Final Answer From the given options, x = − 16 is also an extraneous solution since lo g 7 ( x ) would be lo g 7 ( − 16 ) , which is undefined. However, we are looking for the extraneous solution that we found when solving the equation, which is x = − 4 .
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 or determining the acidity (pH) of a solution in chemistry. Extraneous solutions can arise when manipulating these equations, so it's crucial to check the validity of each solution in the original context. For example, when modeling population growth using logarithmic scales, negative values for population don't make sense, so any negative solutions would be extraneous.
