Question

Log2x(x^2+x-2)=1 ........................

Answer 1

ОДЗ
{x>0
{x≠1/2
{x²+x-2>0⇒x<-2 U x>1
x1+x2=-1 U x1*x2=-2⇒x1=-2 U x2=1
x>1
x∈(1;∞)
x²+x-2=2x
x²-x-2=0
x1+x2=1 U x1*x2=-2
x1=-1∉ОДЗ
х2=2

Answer 2

To solve the equation log2x(x^2 + x - 2) = 1, we can start by simplifying the logarithmic expression:

log2x(x^2 + x - 2) = 1

Using the property of logarithms, we can rewrite this as an exponential equation:

2^1 = x(x^2 + x - 2)

Simplifying further:

2 = x^3 + x^2 - 2x

Rearranging the equation:

x^3 + x^2 - 2x - 2 = 0

Now we need to solve this cubic equation. Unfortunately, there is no general formula to solve cubic equations, but we can try to find rational roots using the rational root theorem and then use numerical methods or calculators to find approximate solutions.

The rational root theorem states that if a rational root (p/q) exists for the equation, it must satisfy the following conditions:

1. p is a factor of the constant term (-2 in this case).
2. q is a factor of the leading coefficient (1 in this case).

The factors of -2 are ±1 and ±2, and the factors of 1 are ±1. So the possible rational roots are: ±1, ±2

By trying these values, we can find that x = 1 is a solution to the equation. Dividing the equation by (x - 1), we get:

(x - 1)(x^2 + 2x + 2) = 0

Now we can solve the quadratic equation x^2 + 2x + 2 = 0. However, this quadratic does not have any real roots, as its discriminant (b^2 - 4ac) is negative. Therefore, the only real solution to the original equation is x = 1.