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

2 Answers
A. S. Adikesavan
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 2x#.

Ratnaker Mehta
Aug 1, 2016

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.

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