Decide the number of solutions of the equation?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:
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.
We notice that 25=5^2!
We'll re-write the equation in this manner:
2*(5^2)^x - 5^x - 1=0
We'll substitute 5^x by another variable, u.
2*u^2 - u - 1=0
We didn't find the values of x, yet!
We'll take logarithms both sides:
ln (5^x)=ln 2
x ln 5= ln 2
The exponential 5^x is always positive, for any value of x!
The equation has just one solution. The only solution is x= ln2/ln5.