Solve for x

(log3x)2 + log3x2 + 1 = 0 

Answer by Edwin McCravy(20056)   (Show Source): You can put this solution on YOUR website!
 

(log3x)2 + log3x2 + 1 = 0

On the second term use the rule

logban = n·logba

(log3x)2 + 2·log3x + 1 = 0

Let the letter w, or any letter other
than x, be such that

w = log3x

Then the above equation becomes

        (w)2 + 2·w + 1 = 0
           w2 + 2w + 1 = 0   
Factoring
        (w + 1)(w + 1) = 0

So the solutions w = -1 and w = -1 are 
equal.

Now since w = log3x, then w = -1 becomes

         log3x = -1

Now use the rule of logarithms that says:

logba = c can be rewritten as a = bc 

to rewrite log3x = -1 as

           x = 3-1

or         x = 1/3

Edwin
  

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