Question

When solving the equation log7 ⁡(x+1) + log7⁡x = log7⁡ 12 a student arrives at a verifiable solution and an extraneous solution. The value of the extraneous solution is

3


-4


-3


5.5

Answer 1

`-4`

Explanation:

The given logarithmic equation is:

`\log_7(x+1)+\log_7x=\log_712`

Recall and apply product rule of logarithms.

`\log_aM+\log_aN=\log_aMN`

We apply this property to the left side of the equation to get;

`\log_7x(x+1)=\log_712`

We take the antilogarithm of both sides to get:

`x(x+1)=12`

We expand to obtain:

`x^2+x=12`

We rewrite in the standard quadratic form:

`x^2+x-12=0`

We factor to obtain:

`(x-3)(x+4)=0`

Either
`(x-3)=0` or
`(x+4)=0`

Either
`x=3` or
`x=-4`

But the domain is
`x\:>\:0`.

Hence
`x=-4` is an extraneous solution.