Question

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

Answer 1

First join the log4 on the left:

log4( x*(x-3) = log4(-7x+21)

Then x = -7, works: -7*(-10)=70 = -7*(-7)+21

x=-3, 18 = 42, does not work

x=3 0=0 works,

However, when one puts x = -7 in the *original* exression, log4(-7) or log4(-10) do not exist (you know why?). So x= -7 is extraneous.

Now x=3 gives log4(0) on the left and right, which does not exist.

So, C is the answer, both are extraneous. Seem to work but indeed don't work in the *original* equation
 

Answer 2

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.