What is the answer to 9^x-3^{x+1}+2=0 ?

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

Solve for x: 9^x-3^{x+1}+2=0

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

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

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

Solution steps

9^{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

e^{e^{x}} = 7

a^{x} = k b^{x}

2^{3 x - 5} = 3^{x + 3}

k, k (x) = x^{6} e^{5 x}

(5 x)^{x} = 5^{5^{5}}

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