Answer:
Option 4th is correct
x = 2
Step-by-step explanation:
Using logarithmic rule:
[tex]\log_b x = \log_b y[/tex]
then; x = y
Given the equation:
[tex]\log_3 (x^2+6x) = \log_3 (2x+12)[/tex]
Apply the rule:
[tex]x^2+6x = 2x+12[/tex]
Subtract 2x from both sides we have;
[tex]x^2+4x =12[/tex]
Subtract 12 from both sides we have;
[tex]x^2+4x-12=0[/tex]
⇒[tex]x^2+6x-2x-12 = 0[/tex]
⇒[tex]x(x+6)-2(x+6)= 0[/tex]
⇒[tex](x+6)(x-2)= 0[/tex]
by zero product property we have:
x+6 =0 and x-2 = 0
⇒x= -6 and x = 2
Since, at x = -6 it does not satisfy the given equation.
Therefore, the only solution to the given equation is, x = 2
