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
t^{2} – 6t + 8 = 0
Factoring quadratic equation, we get
t^{2} – 4t - 2t + 8 = 0
t(t - 4) - 2(t - 4) = 0
(t - 4) (t - 2) = 0
Equating each factor to 0, we get
t = 4 Applying the value of t 2^{x} = 4 2^{x} = 2^{2} x = 2 |
t = 2 Applying the value of t 2^{x} = 2 2^{x} = 2^{1} x = 1 |
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
t^{2} – t - 2 = 0
Factoring quadratic equation, we get
t^{2} – 2t + t - 2 = 0
t (t - 2) + (t - 2) = 0
(t + 1)(t - 2) = 0
Equating each factor to 0, we get
t = -1 Applying the value of t 2^{x} = -1 No solution |
t = 2 Applying the value of t 2^{x} = 2 2^{x} = 2^{1} x = 1 |
So, the value of x is 1.
Problem 3 :
9^{x} – 12(3^{x}) + 27 = 0
Solution :
9^{x} – 12(3^{x}) + 27 = 0
(3^{2})^{x} – 12(3^{x}) + 27 = 0
Let 3^{x} = t
(3^{x})^{2} – 12(3^{x}) + 27 = 0
t^{2} – 12t + 27 = 0
(t - 3) (t - 9) = 0
Equating each factor to 0, we get
t - 3 = 0 t = 3 Applying the value of t 3^{x} = 3^{1} x = 1 |
t - 9 = 0 t = 9 Applying the value of t 3^{x} = 3^{2} x = 2 |
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
(3^{2})^{x} – 3^{x }- 6 = 0
(3^{x})^{2} – 3^{x }- 6 = 0
Let 3^{x }= t
t^{2} – t^{ }- 6 = 0
(t - 3)(t + 2) = 0
t - 3 = 0 t = 3 Applying the value of t 3^{x }= 3 x = 1 |
t + 2 = 0 t = -2 Applying the value of t 3^{x }= -2 No solution |
So, the value of x is 1.
Problem 5 :
25^{x} – 23(5^{x}) - 50 = 0
Solution :
25^{x} – 23(5^{x}) - 50 = 0
(5^{2})^{x} – 23(5^{x}) - 50 = 0
(5^{x})^{2 }– 23(5^{x}) - 50 = 0
Let 5^{x} = t
t^{2 }– 23t - 50 = 0
(t - 25) (t + 2) = 0
Equating each factor to 0, we get
t - 25 = 0 t = 25 Applying the value of t 5^{x} = 25 5^{x} = 5^{2} x = 2 |
t + 2 = 0 t = -2 Applying the value of t 5^{x} = -2 No solution |
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
t^{2} + 8t + 7 = 0
(t + 7)(t + 1) = 0
Equating each factor to 0, we get
t + 7 = 0 t = -7 |
t + 1 = 0 t = -1 |
So, the value of x is no
solutions.
May 21, 24 08:51 PM
May 21, 24 08:51 AM
May 20, 24 10:45 PM