Question

1.Log1/3(x^2+3x-1)=-2

Answer 1

X^2 +3x - 1 = (1/3)^(-2);
x^2 + 3x -1 = 3^2;
x^2 + 3x - 1 = 9;
x^2 + 3x - 10=0;
D = 9+40= 49=7^2;
x1 = -5;
 x2 = 2.
Проверим корни на соответствие одз:
x= -5;   x^2 +3x -1 = 25-15-1 = 9 >0;
x= 2;  x^2 +3x - 1 = 4+6-1 = 9 >0 .
Оба корня подходят. Ответ х=  - 5 или х = 2

Answer 2

To solve the equation log₁/₃(x² + 3x - 1) = -2, we can rewrite it in exponential form.

In general, the logarithmic equation logₐ(b) = c can be expressed as aᶜ = b.

Applying this to our equation, we have ₁/₃^(-2) = x² + 3x - 1.

Simplifying further, we get ₃² = x² + 3x - 1.

Expanding the left side, we have 9 = x² + 3x - 1.

Rearranging the equation, we get x² + 3x - 10 = 0.

Now, we can solve this quadratic equation using factoring, completing the square, or the quadratic formula.

To solve it using factoring, we look for two numbers whose sum is 3 and whose product is -10. The numbers are 5 and -2.

Therefore, the equation factors as (x + 5)(x - 2) = 0.

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

x + 5 = 0 => x = -5 x - 2 = 0 => x = 2

So the solutions to the equation log₁/₃(x² + 3x - 1) = -2 are x = -5 and x = 2.