Question

3^x>1/27 2^x<1/8 (2/5)^x+2 >(2/5)^-1

Answer 1

ОТВЕТ ВО ВЛОЖЕНИИИИИИИ

Answer 2

To solve the inequalities, let's work on each one separately:

1. 3^x > 1/27:

To solve this inequality, we need to find the range of values for x that satisfy the inequality. We can start by expressing both sides with the same base:

1/27 = (1/3)^3

So, the inequality becomes:

3^x > (1/3)^3

Now, we can compare the exponents:

x > 3

Thus, the solution to the first inequality is x > 3.

2. 2^x < 1/8:

Again, let's express both sides with the same base:

1/8 = (1/2)^3

The inequality becomes:

2^x < (1/2)^3

Now, we can compare the exponents:

x < 3

Thus, the solution to the second inequality is x < 3.

3. (2/5)^x + 2 > (2/5)^-1:

First, simplify the right-hand side:

(2/5)^-1 = 5/2

Now, we have the inequality:

(2/5)^x + 2 > 5/2

Subtract 2 from both sides:

(2/5)^x > 5/2 - 2

(2/5)^x > 5/2 - 4/2

(2/5)^x > 1/2

To solve for x, we can take the logarithm (base 2/5) of both sides:

x > log_(2/5)(1/2)

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

x > log(1/2) / log(2/5)

Now, calculate the value:

x > -0.4307...

Thus, the solution to the third inequality is x > -0.4307.

In summary, the solutions to the inequalities are:

1. x > 3
2. x < 3
3. x > -0.4307