What is the answer to 9^x-3^{x+1}+1=0 ?

The answer to 9^x-3^{x+1}+1=0 is x=(ln((3+sqrt(5))/2))/(ln(3)),x=(ln((3-sqrt(5))/2))/(ln(3))

Solve for x: 9^x-3^{x+1}+1=0

The solution is x=(ln((3+sqrt(5))/2))/(ln(3)),x=(ln((3-sqrt(5))/2))/(ln(3))

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

x = \frac{ln ( \frac{3 + 5}{2} )}{ln ( 3 )}, x = \frac{ln ( \frac{3 - 5}{2} )}{ln ( 3 )}

Decimal Notation

x = 0.87603 \dots, x = - 0.87603 \dots

Solution steps

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

Apply exponent rules

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

Rewrite the equation with

3^{x} = u

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

Solve

u^{2} - u \cdot 3^{1} + 1 = 0 : u = \frac{3 + 5}{2}, u = \frac{3 - 5}{2}

u = \frac{3 + 5}{2}, u = \frac{3 - 5}{2}

Substitute back

u = 3^{x},

solve for

x

Solve

3^{x} = \frac{3 + 5}{2} : x = \frac{ln ( \frac{3 + 5}{2} )}{ln ( 3 )}

Solve

3^{x} = \frac{3 - 5}{2} : x = \frac{ln ( \frac{3 - 5}{2} )}{ln ( 3 )}

x = \frac{ln ( \frac{3 + 5}{2} )}{ln ( 3 )}, x = \frac{ln ( \frac{3 - 5}{2} )}{ln ( 3 )}

3^{2 x + 2} = 81

3^{(n - 1)} = 1200

2^{x + 1} + 2^{x} - 2^{x - 1} = 3^{3} - 7

2^{x + 1} = 3 (5^{x})

\frac{1}{e ^{t}} = e^{- t}

What is the answer to 9^x-3^{x+1}+1=0 ?
The answer to 9^x-3^{x+1}+1=0 is x=(ln((3+sqrt(5))/2))/(ln(3)),x=(ln((3-sqrt(5))/2))/(ln(3))
Solve for x: 9^x-3^{x+1}+1=0
The solution is x=(ln((3+sqrt(5))/2))/(ln(3)),x=(ln((3-sqrt(5))/2))/(ln(3))