Question

Which of the following shows the extraneous solution(s) to the logarithmic equation? log4 (x) + log4(x - 3) = log4 (-7x + 21)

A. x= -7
B. x= -3
C. x= 3 and x= -7
D. x=7 and x= -3

Answer 1

Answer: The correct option is C, i.e., x=3 and x=-7.

Explanation:

The given equation is,

$log_4(x)+log_4(x-3)=log_4(-7x+21)$

Using product property of logarithm $log_a(mn)=log_am+log_an$

$log_4(x(x-3))=log_4(-7x+21)$

$log_4(x^2-3x)=log_4(-7x+21)$

On comparing both sides,

$x^2-3x=-7x+21$

$x^2-3x+7x-21=0$

$x^2+4x-21=0$

Using grouping method 4x can be written as (7x-3x). Because the product of 7 and -3 is equal to the constant term -21 and the addition of 7 and -3 is equal to the coefficient of x.

$x^2+7x-3x-21=0$

$x(x+7)-3(x+7)=0$

$(x-3)(x+7)=0$

Equate each factor equal to 0.

The value of x is 3 and -7. Therefore, the option C is correct.

Answer 2

The extraneous solutions are x=3 or x=−7.

We have the equation; log4 (x) + log4(x - 3) = log4 (-7x + 21) we can write;

log4[x(x - 3)] = log4 (-7x + 21)

log4 can be cancelled out from both sides and we have;

x^2 - 3x = -7x + 21

To have a proper quadratic equation;

x^2 + 4x -21 = 0

The solutions of the quadratic equation therefore are x=3 or x=−7.

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