Question

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

2log5(x+1)=2


A) x = -6

B) x = -4

C) x = -2

D) x = -1

Answer 1

its a

Explanation:

Answer 2

A)
`x=-6`

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.

`√((x+1)^2)=√(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.