What is the answer to 2*3^{2x}-5*3^x+2=0 ?

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

Solve for x: 2*3^{2x}-5*3^x+2=0

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

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2 \cdot 3^{2 x} - 5 \cdot 3^{x} + 2 = 0

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

Decimal Notation

x = 0.63092 \dots, x = - 0.63092 \dots

Solution steps

2 \cdot 3^{2 x} - 5 \cdot 3^{x} + 2 = 0

Apply exponent rules

2 (3^{x})^{2} - 5 \cdot 3^{x} + 2 = 0

Rewrite the equation with

3^{x} = u

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

Solve

2 u^{2} - 5 u + 2 = 0 : u = 2, u = \frac{1}{2}

u = 2, u = \frac{1}{2}

Substitute back

u = 3^{x},

solve for

x

Solve

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

Solve

3^{x} = \frac{1}{2} : x = - \frac{ln ( 2 )}{ln ( 3 )}

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

2 e^{x} + 6 e^{- x} = 7

x^{x} = 64

e^{x} (2 e^{x} - 1) = 10

(\frac{1}{36})^{x} = 21 6^{x + 5}

2^{x + 1} = 5^{x}

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