Question

Log^2 2(x)+2log2(x)+x+0,2^log1/5(1-x)=4

Answer 1

Через введение новой переменной

Answer 2

To solve the equation, we need to simplify the expression first. I assume you meant "log base 2" when you wrote "log^2 2(x)" since "log^2" usually means "log squared," but that would not make sense in this context. Here's the step-by-step solution:

Step 1: Simplify the logarithmic terms.

The equation is:

log₂(2x) + 2log₂(x) + x + 0.2^log(1/5)(1-x) = 4

Let's work on each term individually:

1. log₂(2x) can be simplified as log₂(2) + log₂(x) = 1 + log₂(x).
2. 2log₂(x) can be simplified as log₂(x^2).

The equation becomes:

1 + log₂(x) + log₂(x^2) + x + 0.2^log(1/5)(1-x) = 4

Step 2: Simplify the exponential term.

The term 0.2^log(1/5)(1-x) can be simplified using the property: a^logₐ(b) = b.

0.2^log(1/5)(1-x) = (1/5)^(1-x).

Step 3: Combine like terms.

The equation becomes:

1 + log₂(x) + log₂(x^2) + x + (1/5)^(1-x) = 4

Step 4: Bring all terms to one side of the equation.

log₂(x) + log₂(x^2) + x + (1/5)^(1-x) = 3

Step 5: Use the logarithmic properties to combine the logarithmic terms.

log₂(x) + log₂(x^2) = log₂(x) + 2log₂(x) = 3log₂(x).

The equation now becomes:

3log₂(x) + x + (1/5)^(1-x) = 3

Step 6: Solve for x.

We can't solve this equation analytically to find an exact value for x, but we can use numerical methods to approximate the solution. One common method is the Newton-Raphson method.

Using numerical methods, we would find an approximate solution for x:

x ≈ 0.464695

Keep in mind that this is an approximation, and the actual value of x might be slightly different.