Question

How do you solve `log_6 2x + log_6 (x + 3) = 2` and find any extraneous solutions?

Answer 1

`x = 3. x = - 6` is an inadmissible (extraneous ) solution.

Explanation:

Use `log_6 6^2=2 log_6 6=2`

Here, `log_6 2x +log_6(x+3)=log_6( (2x(x+3))=log_6 36`,

So, 2x(x+3)=36. solving,

`x=3, -6`.

Negative x is inadmissible for `log_6 2x`.

Answer 2

The Soln. is `x=3`, while, `x=-6` is extraneous.

Explanation:

We use the following Rules of Logarithm Function :-

`(1)` : Defn. `: log_b y=m iff b^m=y, where, b>0, b!=1, y in RR^+`.

`(2) : log_b a+log_b c=log_b ac`.

Given that,

`log_6 2x+log_6 (x+3)=2`

`rArr log_6 {2x(x+3)}=2`

`rArr 2x(x+3)=6^2=36`

`rArr x(x+3)=18=3(3+3)`

By Inspection , `x=3` is a root. Since, it is a quadr. poly. , it
must have one more root. To find it, we observe that the product
of the roots is `-18`. Hence, the other root must be `-18/3=-6`

But, `x=-6` makes the L.H.S. undefined , and, hence, it is an Extraneous Soln.

`x=3` clearly satisfies the given eqn.

Thus, the soln. is `x=3`, while, `x=-6` is extraneous.