Question

(log2x(4x^3))^2 -2=log2x(4x)

Answer 1

Ответ:

Пошаговое объяснение:

!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Answer 2

Ограничения для х:

4х больше 0   2х больше 0           2х не равно 1

х больше 0                                         х не равно 0,5

Введем замену log2x(4x)=t

Перенесем все в левую часть уравнения

t^2-t-2=0

t1=-1

t2=2

Обратная замена:

log2x(4x)=-1

4x=1/2x

х^2=1/8

х=корень из 2/4

х=-корень из 2/4

log2x(4x)=2

4х=4х^2

x=0

x=1

Answer 3

Let's solve the given equation step-by-step. I'll assume that the expression is as follows:

(log2(x)(4x^3))^2 - 2 = log2(x)(4x)

First, let's simplify the left side of the equation:

Step 1: Simplify the square of the logarithm (log2(x)(4x^3))^2 = (log2(x)(4x^3)) * (log2(x)(4x^3))

Step 2: Apply the property of logarithms: log_b(a^c) = c * log_b(a) (log2(x)(4x^3))^2 = 2 * log2(x)(4x^3)

Now our equation becomes:

2 * log2(x)(4x^3) - 2 = log2(x)(4x)

Next, we'll try to simplify the terms involving logarithms:

Step 3: Apply the property of logarithms: log_b(a * c) = log_b(a) + log_b(c) 2 * log2(x)(4x^3) = log2(x)(4) + log2(x)(x^3)

Step 4: Simplify further: 2 * log2(x)(4x^3) = 2 + 3 * log2(x)(x)

Now the equation becomes:

2 + 3 * log2(x)(x) - 2 = log2(x)(4x)

Step 5: Cancel out the common term "2" on both sides:

3 * log2(x)(x) = log2(x)(4x)

Step 6: Divide both sides by 3:

log2(x)(x) = (1/3) * log2(x)(4x)

Now, we need to remember the property of logarithms that states: log_b(a) = c is equivalent to b^c = a.

Step 7: Rewrite the equation using the property:

x = (4x)^(1/3)

Step 8: Solve for x:

Cube both sides to eliminate the cube root:

x^3 = 4x

Step 9: Rearrange the equation to set it to zero:

x^3 - 4x = 0

Step 10: Factor out the common term "x":

x(x^2 - 4) = 0

Now, we have two potential solutions:

1. x = 0
2. x^2 - 4 = 0

Step 11: Solve for the second solution:

x^2 - 4 = 0

x^2 = 4

x = ±√4

So, the second solution gives us two values: x = 2 and x = -2.

In conclusion, the three solutions to the equation are x = 0, x = 2, and x = -2. Please note that when using logarithms, we need to check that the values of x found are valid within the domain of the original equation. In this case, since we have a logarithm with base 2, x must be greater than 0. Therefore, x = 0 is not a valid solution for the original equation. The valid solutions are x = 2 and x = -2.