Tutor Josmiceli gave these four answers:






His work was correct up to that point.  But he didn't finish.

1. He didn't rule out the two extraneous answers.
2. He didn't simplify the two answers which are solutions.

His second and fourth answers are negative which are not in
the domain of the log function.  So the only solutions are




or 

However they are not in simplest radical form, as they
contain a square root inside a square root.

We set  

 
where u and v are rational.

Square both sides:



We equate the terms without square roots and
equate terms with square roots, and we have this
system:



Divide the second equation through by 



Squaring both sides:





Substituting in





Multiply through by u2







;  
;  
;   

Since we assumed u and v are rational, we discard
the second answers for u.

Therefore:



However we must discard the negative values  and
 as extraneous.

So the only two solutions in simplified form are



Edwin