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

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neela's profile pic

neela | High School Teacher | (Level 3) Valedictorian

Posted on

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.

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

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

u1=1+sqrt (1+4*2)

u1=1+sqrt (9)

u1=1+3

u1=2

u2=1-sqrt (1+4*2)

u2=1-3

u2=-1

We didn't find the values of x, yet!

5^x=u1

5^x=2

We'll take logarithms both sides:

ln (5^x)=ln 2

x ln 5= ln 2

x= ln2/ln5

5^x=u2

5^x=-1 impossible!

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.

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