Question

1/9)^x-6(1/3)^x>-9 решите

Answer 1

Ответ:

ушарко цзаьстеиеок

Объяснение:

епаровадьтимоха

Answer 2

To solve the inequality (1/9)^x - 6(1/3)^x > -9, we can use a substitution to simplify it. Let's make a substitution:

Let y = (1/3)^x

Now, the inequality becomes:

(1/9)^x - 6y > -9

Next, let's rewrite (1/9)^x in terms of y:

(1/9)^x = (1/3)^{2x} = y^2

Now, our inequality becomes:

y^2 - 6y > -9

To solve this quadratic inequality, we need to find the points where it is equal to zero and then check the sign of y in the intervals determined by those points.

Step 1: Find the points where y^2 - 6y = 0 y^2 - 6y = 0 y(y - 6) = 0

So, the critical points are y = 0 and y = 6.

Step 2: Check the sign of y in the intervals determined by the critical points.

Interval 1: y < 0 Choose y = -1 (any value less than 0 will work) (-1)^2 - 6(-1) > -9 1 + 6 > -9 7 > -9 (True)

Interval 2: 0 < y < 6 Choose y = 1 (any value between 0 and 6 will work) 1^2 - 6(1) > -9 1 - 6 > -9 -5 > -9 (True)

Interval 3: y > 6 Choose y = 7 (any value greater than 6 will work) 7^2 - 6(7) > -9 49 - 42 > -9 7 > -9 (True)

Step 3: Analyze the results

The inequality (1/9)^x - 6(1/3)^x > -9 holds true for all values of y in the intervals: y < 0, 0 < y < 6, and y > 6.

Step 4: Convert back to the original variable x

Recall that y = (1/3)^x So, for each interval, we have:

Interval 1: (1/3)^x < 0 However, (1/3)^x can never be negative since it is always positive (a positive number raised to any power is positive).

Interval 2: 0 < (1/3)^x < 6 Now, take the logarithm base 3 on both sides: x log(1/3) < log(6) x < log(6) / log(1/3)

Interval 3: (1/3)^x > 6 Take the logarithm base 3 on both sides: x log(1/3) > log(6) x > log(6) / log(1/3)

Final Solution:

The solution to the inequality is: x < log(6) / log(1/3) for 0 < y < 6 x > log(6) / log(1/3) for y > 6

Approximating the value of log(6) / log(1/3), we get: x < -2.585 or x > 7.292

So, the solution for x is: x < -2.585 or x > 7.292