Question

Log^2(0.5,x)+log(0.5,x)-2=0

Answer 1

Ответ:

$log0,5(x)=t\\t^2-5+2=0\\t=-2\\t=1\\log0,5(x)=-2\\x=4\\log0,5(x)=1\\x=0,5$

Пошаговое объяснение:

Answer 2

To solve the equation log^2(0.5, x) + log(0.5, x) - 2 = 0, we can use a substitution to simplify the equation.

Let's substitute log(0.5, x) with a variable, say "y."

Substituting, the equation becomes:

y^2 + y - 2 = 0.

Now, we have a quadratic equation. To solve it, we can factor it or use the quadratic formula.

Factoring the equation, we have:

(y + 2)(y - 1) = 0.

Setting each factor equal to zero, we get two possible solutions:

1. y + 2 = 0 y = -2

2. y - 1 = 0 y = 1

Now, substitute back the value of y into the original substitution:

1. For y = -2: log(0.5, x) = -2

   Since the logarithm of a positive number cannot be negative, there are no solutions for this case.

2. For y = 1: log(0.5, x) = 1

   We can rewrite this equation as an exponential equation:

   0.5^1 = x 0.5 = x

So the solution to the equation log^2(0.5, x) + log(0.5, x) - 2 = 0 is x = 0.5.