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

Aug 1, 2016

$x = 3. x = - 6$ 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 ${\log}_{6} 2 x$.

Aug 1, 2016

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

Explanation:

We use the following Rules of Logarithm Function :-

$\left(1\right)$ : Defn. $: {\log}_{b} y = m \iff {b}^{m} = y , w h e r e , b > 0 , b \ne 1 , y \in {\mathbb{R}}^{+}$.

$\left(2\right) : {\log}_{b} a + {\log}_{b} c = {\log}_{b} a c$.

Given that,

${\log}_{6} 2 x + {\log}_{6} \left(x + 3\right) = 2$

$\Rightarrow {\log}_{6} \left\{2 x \left(x + 3\right)\right\} = 2$

$\Rightarrow 2 x \left(x + 3\right) = {6}^{2} = 36$

$\Rightarrow x \left(x + 3\right) = 18 = 3 \left(3 + 3\right)$

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 $- \frac{18}{3} = - 6$

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.