Decide the number of solutions of the equation?2*25^x = 1+5^x

2*25^x =1+5^x

To solve this
equation, we put 5^x = y.

Then 25^x = (5^2)x = (5^x)^2 =
y^2.

So now we can rewrite the given equation
as:

2*y^2 =1+y

2y^2 -y -1 =
0

(2y+1)(y-1) = 0

2y+1 = 0 or
y-1 = 0.

y -1 = 0 gives: y = 1.
Or

5^x = 1 5^0

x =
0.

2y+1  = 0 gives: y = -1/2..
Or

5^x = -1/2. Thereis nor real solution. as for all real
x, 5^x > 0.

Therefore there is only one real
solution , that is , x = 0.