Question

Lg^2(x/10)+lg(10x)>2

Answer 1

Ответ:

Объяснение:

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Answer 2

To solve the inequality lg^2(x/10) + lg(10x) > 2, let's break it down step-by-step.

First, let's rewrite the logarithms using common properties of logarithms:

1. lg^2(x/10) = (lg(x/10))^2
2. lg(10x) = lg(10) + lg(x) = 1 + lg(x) (since lg(10) = 1)

Now, our inequality becomes:

(lg(x/10))^2 + 1 + lg(x) > 2

Subtract 2 from both sides to move everything to the left side:

(lg(x/10))^2 + lg(x) - 1 > 0

Now, let u = lg(x/10), so we have:

u^2 + lg(10^u) - 1 > 0

The inequality now becomes:

u^2 + u - 1 > 0

Now, we need to find the solutions to this quadratic inequality. To do that, let's find the roots of the equation u^2 + u - 1 = 0:

Using the quadratic formula (or factoring), the roots are:

u = (-1 ± √5) / 2

Approximately, u ≈ 0.61803 and u ≈ -1.61803

Now, we need to determine the sign of the expression u^2 + u - 1 for different intervals.

1. When u < -1.61803: u^2 + u - 1 is positive since both u^2 and u are positive, and -1 is negative.

2. When -1.61803 < u < 0.61803: u^2 + u - 1 is negative since u^2 is positive, u is negative, and -1 is negative.

3. When u > 0.61803: u^2 + u - 1 is positive since both u^2 and u are positive, and -1 is negative.

Now, we need to translate the values of u back into x values using u = lg(x/10):

1. When lg(x/10) < -1.61803: x/10 < 10^(-1.61803)

2. When -1.61803 < lg(x/10) < 0.61803: 10^(-1.61803) < x/10 < 10^(0.61803)

3. When lg(x/10) > 0.61803: x/10 > 10^(0.61803)

Finally, we can convert the above inequalities back to x:

1. x < 10^(-1.61803 + 1) = 10^(-0.61803) ≈ 0.2367

2. 10^(-1.61803) < x < 10^(0.61803 + 1) = 10^(1.61803) ≈ 5.0589

3. x > 10^(0.61803 + 1) = 10^(1.61803) ≈ 5.0589

So, the solution to the inequality is approximately:

x < 0.2367 or x > 5.0589

Keep in mind that logarithms are not defined for non-positive values, so x must be greater than 0 for this inequality to be valid. Therefore, the final solution is:

x ∈ (0, 0.2367) ∪ (5.0589, +∞)