Question

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

Answer 1

Answer:

A) $x=-6$

Step-by-step explanation:

$2\log_5 (x+1)=2$

We can use logarithm properties to write it as:

$\log_5 (x+1)^2=2$

On removing $log$ it gives:

$(x+1)^2=5^2$

Taking square root both sides.

$\sqrt{(x+1)^2}=\sqrt{5^2}$

$(x+1)=\pm 5$

Solving for $x$ we get:

$x+1=5$ and $x+1=-5$

Subtracting by 1 to both sides of both the above equations.

$x+1-1=5-1$ and $x+1-1=-5-1$

$x=4$ and $x=-6$

So we have got two solutions for the logarithmic equation.

Now we check the validity of the solutions by plugging values in the equation $2\log_5 (x+1)=2$

Plugging $x=4$

$2\log_5 (4+1)=2$

$2\log_5 5=2$

$2(1)=2$  [As $log_5 5=1$]

$2=2$

So, thats a valid solution.

Plugging $x=-6$

$2\log_5 (-6+1)=2$

$2\log_5 (-5)=2$

As $\log_5 (-5)$ is undefined, so $x=-6$ is not a valid solution which means its extraneous.

Answer 2

Answer:

its a

Step-by-step explanation: