SOLVING QUADRATIC POLYNOMIALS WITH VARIABLE EXPONENTS

Solve for x.

Problem 1 :

4^x – 6(2^x) + 8 = 0

Solution :

4^x – 6(2^x) + 8 = 0

(2^x)² – 6(2^x) + 8 = 0

Let 2^x =  t

t² – 6t + 8 = 0

Factoring quadratic equation, we get

t² – 4t - 2t + 8 = 0

t(t - 4) - 2(t - 4) = 0

(t - 4) (t - 2) = 0

Equating each factor to 0, we get

So, the values of x is 1 or 2.

Problem 2 :

4^x – 2^x - 2 = 0

Solution :

4^x - 2^x – 2 = 0

(2^x)² – 2^x - 2 = 0

Let 2^x =  t

t² – t - 2 = 0

Factoring quadratic equation, we get

t² – 2t + t - 2 = 0

t (t - 2) + (t - 2) = 0

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

Equating each factor to 0, we get

So, the value of x is 1.

Problem 3 :

9^x – 12(3^x) + 27 = 0

Solution :

9^x – 12(3^x) + 27 = 0

(3²)^x – 12(3^x) + 27 = 0

Let 3^x = t

(3^x)² – 12(3^x) + 27 = 0

t² – 12t + 27 = 0

(t - 3) (t - 9) = 0

Equating each factor to 0, we get

So, the values of x is 1 or 2.

Problem 4 :

9^x = 3^x + 6

Solution :

9^x = 3^x + 6

9^x - 3^x – 6 = 0

(3²)^x – 3^x - 6 = 0

(3^x)² – 3^x - 6 = 0

Let 3^x = t

t² – t - 6 = 0

(t - 3)(t + 2) = 0

So, the value of x is 1.

Problem 5 :

25^x – 23(5^x) - 50 = 0

Solution :

25^x – 23(5^x) - 50 = 0

(5²)^x – 23(5^x) - 50 = 0

(5^x)² – 23(5^x) - 50 = 0

Let 5^x = t

t² – 23t - 50 = 0

(t - 25) (t + 2) = 0

Equating each factor to 0, we get

So, the value of x is 2.

Problem 6 :

49^x + 8(7^x) + 7 = 0

Solution :

49^x + 8(7^x) + 7 = 0

(7^x)² + 8(7^x) + 7 = 0

Let 7^x = t

t² + 8t + 7 = 0

(t + 7)(t + 1) = 0

Equating each factor to 0, we get

So, the value of x is no solutions.