Question

9^x-4×3^(x+1)+27=0 найдите x​

Answer 1

$9^{x}-4*3^{x+1}+27=0\\\\(3^{x})^{2}-4*3^{x}*3+27=0\\\\(3^{x} )^{2}-12*3^{x}+27=0\\\\3^{x}=m,m>0\\\\m^{2} -12m+27=0\\\\m_{1}=3\\\\m_{2}=9-teorema.Vieta\\\\3^{x}=3\\\\x_{1}=1\\\\3^{x}=9\\\\3^{x}=3^{2}\\\\x_{2}=2\\\\Otvet:\boxed{1;2}$

Answer 2

We can simplify the equation by noticing that 27 is equal to 3 raised to the power of 3. Therefore, we can write the equation as:

9^x - 4 × 3^(x+1) + 3^3 = 0

Now we can use the fact that 3^(x+1) is equal to 3 × 3^x, and substitute to get:

9^x - 4 × 3 × 3^x + 27 = 0

We can further simplify by factoring out 3^x:

(9^x - 4 × 3^x × 3) + 27 = 0

(9^x - 12 × 3^x) + 27 = 0

9^x - 12 × 3^x = -27

Dividing both sides by 3^x, we get:

(9/3)^x - 12 = -27/3^x

3^x - 12 = -9/3^x

Multiplying both sides by 3^x, we get:

3^(2x) - 12 × 3^x + 9 = 0

We can now substitute y = 3^x to get a quadratic equation in y:

y^2 - 12y + 9 = 0

Using the quadratic formula, we get:

y = (12 ± √(144 - 4 × 9)) / 2

y = 6 ± 3√3

Substituting back to get x, we get:

x = log_3(6 + 3√3) or x = log_3(6 - 3√3)

Therefore, the solutions are x = log_3(6 + 3√3) or x = log_3(6 - 3√3).