Question

What are the potential solutions of log4x+log4(x+6)=2

Answer 1

Log4x+log4(x+6)=2  x=2

log4(x(x+6))=2                           2=log4(16)
log4(x(x+6))=log4(16)
then you can remove the logs since they are the same
(x(x+6))=16
x^2+6x=16
x^2+6x-16
factor
(x+8)(x-2)
meaning that x= -8,2
but because of log rules you cant have negative logs. so x=2

Answer 2

Answer:  The correct option is

(C) x = 2  and  x = -8.

Step-by-step explanation:  We are given to find the potential solutions of the following logarithmic equation :

$log_4x+\log_4(x+6)=2~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)$

We will be using the following logarithmic properties to solve the given equation :

$(i)~\log_ab+\log_ac=\logabc,\\\\(ii)~\log_ab=c~~~~~\Rightarrow b=a^c.$

The solution of equation (i) is as follows :

$log_4x+\log_4(x+6)=2\\\\\Rightarrow \log_4x(x+6)=2~~~~~~~~~~~~~~~~~~~~[\textup{Using property (i)}]\\\\\Rightarrow x(x+6)=4^2~~~~~~~~~~~~~~~~~~~~~~~[\textup{Using property (ii)}]\\\\\Rightarrow x^2+6x-16=0\\\\\Rightarrow x^2+8x-2x-16=0\\\\\Rightarrow x(x+8)-2(x+8)=\\\\\Rightarrow (x-2)(x+8)=0\\\\\Rightarrow x-2=0,~~~~~x+8=\\\\\Rightarrow x=2,~~-8.$

Thus, the required potential solutions of the given equation are

x = 2  and  x = -8.

Option (C) is CORRECT.