What is the answer to 2^{2x}-5(2^x)+6=0 ?

The answer to 2^{2x}-5(2^x)+6=0 is x=(ln(3))/(ln(2)),x=1

Solve for x: 2^{2x}-5(2^x)+6=0

The solution is x=(ln(3))/(ln(2)),x=1

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2^{2 x} - 5 (2^{x}) + 6 = 0

x = \frac{ln ( 3 )}{ln ( 2 )}, x = 1

Decimal Notation

x = 1.58496 \dots, x = 1

Solution steps

2^{2 x} - 5 (2^{x}) + 6 = 0

Apply exponent rules

(2^{x})^{2} - 5 \cdot 2^{x} + 6 = 0

Rewrite the equation with

2^{x} = u

(u)^{2} - 5 u + 6 = 0

Solve

u^{2} - 5 u + 6 = 0 : u = 3, u = 2

u = 3, u = 2

Substitute back

u = 2^{x},

solve for

x

Solve

2^{x} = 3 : x = \frac{ln ( 3 )}{ln ( 2 )}

Solve

2^{x} = 2 : x = 1

x = \frac{ln ( 3 )}{ln ( 2 )}, x = 1

a^{a} = 2

e^{3 x} = 14

x, f = - 2 ((- 6)^{(x + 5)})

e^{3 x} = 22

e^{3 x} = 75

What is the answer to 2^{2x}-5(2^x)+6=0 ?
The answer to 2^{2x}-5(2^x)+6=0 is x=(ln(3))/(ln(2)),x=1
Solve for x: 2^{2x}-5(2^x)+6=0
The solution is x=(ln(3))/(ln(2)),x=1