What is the answer to 9^x-4(3^{x+1})+27=0 ?

The answer to 9^x-4(3^{x+1})+27=0 is x=2,x=1

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9^x-4(3^{x+1})+27=0

9^{x} - 4 (3^{x + 1}) + 27 = 0

x = 2, x = 1

Solution steps

9^{x} - 4 (3^{x + 1}) + 27 = 0

Apply exponent rules

(3^{x})^{2} - 4 \cdot 3^{x} \cdot 3^{1} + 27 = 0

Rewrite the equation with

3^{x} = u

(u)^{2} - 4 u \cdot 3^{1} + 27 = 0

Solve

u^{2} - 4 u \cdot 3^{1} + 27 = 0 : u = 9, u = 3

u = 9, u = 3

Substitute back

u = 3^{x},

solve for

x

Solve

3^{x} = 9 : x = 2

Solve

3^{x} = 3 : x = 1

x = 2, x = 1

2 = 1. 1^{x}

(1.045)^{x} = 2

3^{x - 1} = \frac{1}{81}

1 0^{1 - x} = 6^{x}

2433 = 3^{a}

What is the answer to 9^x-4(3^{x+1})+27=0 ?
The answer to 9^x-4(3^{x+1})+27=0 is x=2,x=1
Solve for x: 9^x-4(3^{x+1})+27=0
The solution is x=2,x=1