equation

Solve for x over the real numbers:
4-5 2^x+2^(2 x) = 0

 

Simplify and substitute y = 2^x:
 4-5 2^x+2^(2 x) = 4-5×2^x+(2^x)^2 = y^2-5 y+4 = 0:
y^2-5 y+4 = 0

 

The left hand side factors into a product with two terms:
(y-4) (y-1) = 0

 

Split into two equations:
y-4 = 0 or y-1 = 0

 

Add 4 to both sides:
y = 4 or y-1 = 0

 

Substitute back for y = 2^x:
2^x = 4 or y-1 = 0

 

4 = 2^2:
2^x = 2^2 or y-1 = 0

 

Equate exponents of 2 on both sides:
x = 2 or y-1 = 0

 

Add 1 to both sides:
x = 2 or y = 1

 

Substitute back for y = 2^x:
x = 2 or 2^x = 1

 

Express both sides with bases 2:
x = 2 or 2^x = 2^0

 

Equate exponents of 2 on both sides:
Answer: |x = 2     or     x = 0