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

Algebra ->  Exponential-and-logarithmic-functions -> 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      Log On


   

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) About Me  (Show Source):
You can put this solution on YOUR website!





















John

My calculator said it, I believe it, that settles it

Answer by ikleyn(52776) About Me  (Show Source):
You can put this solution on YOUR website!
.

(1)  Introduce new variable  u = 5%5Ex.


     Then the equation becomes

         u%5E2+-+26u+%2B+25 = 0

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


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


    The root  u = 1  means that  5%5Ex = 1,  i.e.  x = 0.


    The root  u = 25   means that  5%5Ex = 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)   2%5Ex = 0.125 = 1%2F8.


      x = -3.    ANSWEWR





(4)  25%5E%285x%29 = 625


     5%5E%2810x%29 = 5%5E4


     10x = 4


       x = 4%2F10 = 0.4.    ANSWER



Answer by Boreal(15235) About Me  (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