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

Algebra ->  Exponential-and-logarithmic-functions -> SOLUTION: \(5^{(}logx-1)/125=(1/5)^{(}logx)^{2}-logx\)      Log On


   

Question 1207490: \(5^{(}logx-1)/125=(1/5)^{(}logx)^{2}-logx\)
Found 2 solutions by ikleyn, Edwin McCravy:
Answer by ikleyn(52790) About Me  (Show Source):
You can put this solution on YOUR website!
.

    %285%5E%28log%28%28x%29%29-1%29%29%2F125 =  %281%2F5%29%5E%28log%28%28x%29%29%5E2-log%28%28x%29%29%29


Write right part with the base 5


    %285%5E%28log%28%28x%29%29-1%29%29%2F125 =  5%5E%28log%28%28x%29%29-log%28%28x%29%29%5E2%29


Simplify left side

    5%5E%28log%28%28x%29%29-4%29 = 5%5E%28log%28%28x%29%29-log%28%28x%29%29%5E2%29.


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) = +/- sqrt%284%29

      log(x) = +/- 2


There are two solutions:  x= 10%5E2 = 100  and  x= 10%5E%28-2%29 = 0.01.    ANSWER

Solved.



Answer by Edwin McCravy(20056) About Me  (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