SOLUTION: Solve the following exponential equation (1)5^2x-26(5^x)+25=0 (2)2^2x+2^x+1-8=0 (3)2^x=0.125 (4)25^5x=625

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Question 1162871: Solve the following exponential equation
(1)5^2x-26(5^x)+25=0
(2)2^2x+2^x+1-8=0
(3)2^x=0.125
(4)25^5x=625
Found 3 solutions by solver91311, ikleyn, Boreal:
Answer by solver91311(24713)   (Show Source): You can put this solution on YOUR website!





















John

My calculator said it, I believe it, that settles it
Answer by ikleyn(52777)   (Show Source): You can put this solution on YOUR website!
.

(1)  Introduce new variable  u = .


     Then the equation becomes

          = 0

         (u-1)*(u-25) = 0.


    It has two roots  u = 1  and  u = 25.


    The root  u = 1  means that   = 1,  i.e.  x = 0.


    The root  u = 25   means that   = 25,  i.e.  x = 2.


    ANSWER.  There are two solutions  0 and 2.




(2)  This equation is presented in the STARNGE form in the post.

   
     I do not believe it is correct.


     Double check it (!)





(3)    = 0.125 = .


      x = -3.    ANSWEWR





(4)   = 625


      = 


     10x = 4


       x =  = 0.4.    ANSWER



Answer by Boreal(15235)   (Show Source): You can put this solution on YOUR website!
1. (5^x-25)*(5^x-1)=0
5^x=25
x=2
5^x=1
x=0
-
2.(2^x-2)(2^x+4)=0
2^x=2
x=1
2^x=-4 no solution
-
3. 2^x=2^(-3)
x=-3
-
4.25^5x=25^2
5x=2
x=0.4
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