What is the answer to 9^x-3^{x+2}+18=0 ?

The answer to 9^x-3^{x+2}+18=0 is x=(ln(6))/(ln(3)),x=1

Solve for x: 9^x-3^{x+2}+18=0

The solution is x=(ln(6))/(ln(3)),x=1

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9^{x} - 3^{x + 2} + 18 = 0

x = \frac{ln ( 6 )}{ln ( 3 )}, x = 1

Solution steps

9^{x} - 3^{x + 2} + 18 = 0

Apply exponent rules

(3^{x})^{2} - 3^{x} \cdot 3^{2} + 18 = 0

Rewrite the equation with

3^{x} = u

(u)^{2} - u \cdot 3^{2} + 18 = 0

Solve

u^{2} - u \cdot 3^{2} + 18 = 0 : u = 6, u = 3

u = 6, u = 3

Substitute back

u = 3^{x},

solve for

x

Solve

3^{x} = 6 : x = \frac{ln ( 6 )}{ln ( 3 )}

Solve

3^{x} = 3 : x = 1

x = \frac{ln ( 6 )}{ln ( 3 )}, x = 1

1.4 \cdot 0. 9^{x} = 0. 5^{x}

10 0^{\frac{2}{x}} = (1 + 0.1)^{x}

2 (1 0^{x}) + 35 = 179

30 = 286.93 (0.973 8^{x})

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

What is the answer to 9^x-3^{x+2}+18=0 ?
The answer to 9^x-3^{x+2}+18=0 is x=(ln(6))/(ln(3)),x=1
Solve for x: 9^x-3^{x+2}+18=0
The solution is x=(ln(6))/(ln(3)),x=1