Question

Which of the following shows the extraneous solution to the logarithmic equation below???

Answer 1

Short Answer x = - 6
Remark
You gave the second best answer. 

The trick here is not to divide both sides by 2. Solve the problem this way.

log_5(x + 1)^2 = 2 Take the antilog of both sides
(x + 1)^2 = 5^2       Expand the equation
x^2 + 2x + 1 = 25   Subtract 25 from both sides.
x^2 + 2x - 24 = 0    Factor
(x + 6)(x - 4) = 0     Find the zeros.

x + 6 = 0
x = - 6  <<<<<<<< Answer. This is the extraneous root.
The reason this is an extraneous root is that x<=0 do not have a logarithem

x - 4 = 0
x = 4    This is a legitimate result to the original equation.

Answer 2

Answer:

B

Step-by-step explanation: