Question

(0.3)^x^2-2x+2=0.09 решите пж!

Answer 1

$(0.3)^{x^2-2x+2}=(0.3)^2 \\ x^2-2x+2=2 \\ x^2-2x=0 \\ x=0|||or|||x=2$

Answer 2

0,3х*2-2х+2=0,09
0,6-2х+2=0,09
-1,4х+2=0,09
-1,4х=0,09-2
-1,4х=-1,91
Дальше мы делим обе стороны уравнения на -1,4
х= 191/140 или 1 51/140

Answer 3

To solve the equation (0.3)^x^2 - 2x + 2 = 0.09, we need to find the values of x that satisfy the equation. Let's proceed step by step:

Step 1: Rewrite 0.09 as (0.3)^2 since 0.3^2 = 0.09.

The equation now becomes:

(0.3)^x^2 - 2x + 2 = (0.3)^2

Step 2: Simplify the equation:

(0.3)^x^2 - 2x + 2 = 0.3^2

Step 3: Subtract 0.3^2 from both sides:

(0.3)^x^2 - 2x + 2 - 0.3^2 = 0

Now, the equation becomes:

(0.3)^x^2 - 2x + 1.91 = 0

Step 4: Unfortunately, this equation does not have a straightforward algebraic solution. However, we can use numerical methods to approximate the solutions. One common numerical method is the Newton-Raphson method.

Without a specific range for x, it's challenging to find the exact solutions. Let's solve for the first few solutions using the Newton-Raphson method, starting from initial guesses for x:

1. Initial guess: x = 0
2. Initial guess: x = 1
3. Initial guess: x = -1

We'll use these initial guesses to iteratively find the solutions:

Iteration 1:

• For x = 0: (0.3)^0^2 - 2(0) + 1.91 ≈ 1.91
• For x = 1: (0.3)^1^2 - 2(1) + 1.91 ≈ 1.51
• For x = -1: (0.3)^(-1)^2 - 2(-1) + 1.91 ≈ 2.91

Iteration 2:

• For x = 0: (0.3)^0^2 - 2(0) + 1.91 ≈ 1.91
• For x = 1: (0.3)^1^2 - 2(1) + 1.91 ≈ 1.51
• For x = -1: (0.3)^(-1)^2 - 2(-1) + 1.91 ≈ 2.91

Since the values do not seem to converge to zero, the initial guesses may not be good approximations. We might need to try different initial guesses or use a more sophisticated numerical method.

Keep in mind that these approximations might not be entirely accurate, and finding exact solutions may be challenging without additional context or constraints on the range of x. If you have a specific range or context for x, please provide it, and we can attempt to find a better solution.