What is the answer to 3^{2x}-3^{x+1}+2=0 ?

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

Solve for x: 3^{2x}-3^{x+1}+2=0

The solution is x=(ln(2))/(ln(3)),x=0

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3^{2 x} - 3^{x + 1} + 2 = 0

x = \frac{ln ( 2 )}{ln ( 3 )}, x = 0

Decimal Notation

x = 0.63092 \dots, x = 0

Solution steps

3^{2 x} - 3^{x + 1} + 2 = 0

Apply exponent rules

(3^{x})^{2} - 3^{x} \cdot 3^{1} + 2 = 0

Rewrite the equation with

3^{x} = u

(u)^{2} - u \cdot 3^{1} + 2 = 0

Solve

u^{2} - u \cdot 3^{1} + 2 = 0 : u = 2, u = 1

u = 2, u = 1

Substitute back

u = 3^{x},

solve for

x

Solve

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

Solve

3^{x} = 1 : x = 0

x = \frac{ln ( 2 )}{ln ( 3 )}, x = 0

2^{3 x + 4} = 4^{2 x + 3}

e^{3 x} = 9

4 = e^{x}

3^{9 - 3 x} = 27

7^{x} = 13

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