Question

8^x - 3*2^(x+2)+2^(5-x) = 0 Найдите х

Answer 1

Вот так выгледит задача ?
если да вот ответ

Answer 2

8^x-3*2^(x+2)+2^(5-x)=0;|:8^x>0

1-3*2^(x+2-3x)+2^(5-x-3x)=0

1-3*2^(2-2x)+2^(5-4x)=0

1-3*4*2^(-2x)+32(*2^(-2x))²=0
2^(-2x)=t>0
32t²-12t+1=0
D/4=36-32=4=2²
t=(6±2)/32
t1=8/32=1/4
t2=4/32=1/8

1)2^(-2x)=1/4
2^(-2x)=2^(-2)
-2x=-2
x=1

2)2^(-2x)=2^(-3)
-2x=-3
x=3/2

ответ 1;3/2

Answer 3

To solve the equation 8^x - 3*2^(x+2) + 2^(5-x) = 0, we need to find the value of x that satisfies the equation. Unfortunately, this equation cannot be solved analytically using simple algebraic methods. However, we can use numerical methods to find an approximate solution.

One common numerical method for solving equations is the Newton-Raphson method. Let's use this method to find an approximate value for x.

1. Start with an initial guess for x (let's call it x_0).
2. Use the formula: x_(n+1) = x_n - f(x_n) / f'(x_n), where f(x) is the equation we want to solve, and f'(x) is its derivative.
3. Repeat the process, plugging in x_n to find x_(n+1) until the difference between consecutive approximations is small enough.

To make the calculation a bit easier, let's rewrite the equation as follows: 8^x - 32^(x+2) + 2^(5-x) = 0 (2^3)^x - 32^(x+2) + 2^(5-x) = 0 2^(3x) - 3*2^(x+2) + 2^(5-x) = 0

Now, let's find the derivative of the equation with respect to x: f(x) = 2^(3x) - 32^(x+2) + 2^(5-x) f'(x) = d/dx (2^(3x)) - d/dx (32^(x+2)) + d/dx (2^(5-x))

To find the derivative of 2^(3x), we use the chain rule: d/dx (2^(3x)) = ln(2) * 2^(3x) * 3 = 3ln(2) * 2^(3x)

To find the derivative of 32^(x+2), we use the chain rule as well: d/dx (32^(x+2)) = ln(2) * 3 * 2^(x+2) = 3ln(2) * 2^(x+2)

To find the derivative of 2^(5-x), we use the chain rule: d/dx (2^(5-x)) = ln(2) * 2^(5-x) * (-1) = -ln(2) * 2^(5-x)

Now, we can rewrite the equation and its derivative: Equation: 2^(3x) - 3*2^(x+2) + 2^(5-x) = 0 Derivative: f'(x) = 3ln(2) * 2^(3x) - 3ln(2) * 2^(x+2) - ln(2) * 2^(5-x)

Let's apply the Newton-Raphson method:

Step 1: Choose an initial guess for x (x_0). Let's use x_0 = 1.

Step 2: Calculate x_1: x_1 = x_0 - f(x_0) / f'(x_0)

Step 3: Repeat Step 2 until convergence.

I'll perform the calculations for the first iteration:

x_0 = 1 f(x_0) = 2^(31) - 32^(1+2) + 2^(5-1) = 8 - 38 + 2^4 = 8 - 24 + 16 = 0 f'(x_0) = 3ln(2) * 2^(31) - 3ln(2) * 2^(1+2) - ln(2) * 2^(5-1) = 3ln(2) * 8 - 3ln(2) * 8 - ln(2) * 2^4 = 3ln(2)*8 - 3ln(2)*8 - ln(2)*16 = -8ln(2)

x_1 = x_0 - f(x_0) / f'(x_0) = 1 - 0 / (-8ln(2)) = 1

Since x_1 is the same as x_0, it means the method has converged to an approximate solution.

Therefore, the value of x that satisfies the equation is approximately x = 1. Keep in mind that this is just an approximate solution obtained using numerical methods, and the exact solution may not have a simple algebraic representation.