SOLUTION: \(5^{(}logx-1)/125=(1/5)^{(}logx)^{2}-logx\)

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Question 1207490: \(5^{(}logx-1)/125=(1/5)^{(}logx)^{2}-logx\)
Found 2 solutions by ikleyn, Edwin McCravy:
Answer by ikleyn(52794)   (Show Source): You can put this solution on YOUR website!
.

     =  


Write right part with the base 5


     =  


Simplify left side

     = .


Since the bases are equal, it implies that the indexes are equal, too

    log(x)-4 = log(x) - (log(x))^2


Simplify

     -4 = -(log(x))^2

      4 = (log(x))^2


Take square root of both sides

      log(x) = +/- 

      log(x) = +/- 2


There are two solutions:  x=  = 100  and  x=  = 0.01.    ANSWER

Solved.



Answer by Edwin McCravy(20056)   (Show Source): You can put this solution on YOUR website!
I was just going to post Ikleyn's solution before she posted it. I had tried
another version of what you notation might mean and finally stumbled onto
the same one she interpreted it to be.  I am unfamiliar with that notation as
things like {(} and backward slashes \ are foreign to me.  What notation is that? 
I've seen it elsewhere.
  
Edwin
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