Hi,

 +  == 1

 3^1 = (x-4)(x-2)

   x^2 - 6x + 5 = 0

  (x - 5)(x-1) = 0   

         x = 5     (x = 1  is an extraneous solution)

and 

 

Answer by solver91311(24713)   (Show Source): You can put this solution on YOUR website!
 





So near and yet so far.  Pay more attention to your signs:











So your quadratic becomes











Which factors tidily, of course.  But you have to be careful with the log function where the domain is restricted to positive real numbers.  One of the roots of your quadratic is good, but the other makes the argument of both log functions less than zero.  Hence this root must be excluded from the solution set of your equation.





John



My calculator said it, I believe it, that settles it



 

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