SOLUTION: : I need help solving for x algebraically. (log base 3(x-4))+(log base 3 (x-2)) =1 I am so lost. I thought I could cancel out logs ad get x^2 -2x -4x -8 = 3^1 x^2 -6x -11 =0 h

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: : I need help solving for x algebraically. (log base 3(x-4))+(log base 3 (x-2)) =1 I am so lost. I thought I could cancel out logs ad get x^2 -2x -4x -8 = 3^1 x^2 -6x -11 =0 h      Log On


   

Question 634223: : I need help solving for x algebraically.
(log base 3(x-4))+(log base 3 (x-2)) =1
I am so lost. I thought I could cancel out logs ad get
x^2 -2x -4x -8 = 3^1
x^2 -6x -11 =0
hm.... help?
Found 2 solutions by ewatrrr, solver91311:
Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!
 

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) About Me  (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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