Question

How many pair of numbers (x;y) that satisfy

log₃ (x² + y² + x) + log₂ (x² + y²) ≤ log₃ (x) + log₂ (x² + y² + 24x)

Answer 1

We can simplify the given inequality using the properties of logarithms:

log₃ (x² + y² + x) + log₂ (x² + y²) ≤ log₃ (x) + log₂ (x² + y² + 24x)

log₃ [(x² + y² + x) / x] + log₂ [(x² + y²) / (x² + y² + 24x)] ≤ 0

log₃ [(x + y²) / x] + log₂ [(1 - 24x / (x² + y² + 24x))] ≤ 0

log₃ [(x + y²) / x] + log₂ [(x² + y² + 24x - 24x) / (x² + y² + 24x))] ≤ 0

log₃ [(x + y²) / x] + log₂ [(x² + y²) / (x² + y² + 24x))] ≤ 0

log₃ [(x + y²) / x] - log₂ [(x² + y² + 24x) / (x² + y²))] ≥ 0

log₃ [(x + y²) / x] - log₂ [(x + 24) / x] ≥ 0

log₃ [(x + y²) / x] ≥ log₂ [(x + 24) / x]

(x + y²) / x ≥ (x + 24) / x^2

x + y² ≥ x + 24

y² ≥ 24

y ≤ ± 2√6

Therefore, the system of inequalities that satisfies the given inequality is:

y ≤ 2√6, y ≥ -2√6

For each value of y between -2√6 and 2√6, there is a corresponding range of x values that satisfies the inequality.

For example, if y = 0, then the inequality simplifies to:

log₃ (x) + log₂ (x²) ≤ 0

log₃ x + 2 log₂ x ≤ 0

log₃ x + log₂ x² ≤ 0

log₆ x³ ≤ 0

x³ ≤ 1

x ≤ 1

So, if y = 0, then the possible values of x are:

0 < x ≤ 1

Thus, for each value of y between -2√6 and 2√6, there is a corresponding range of x values that satisfies the inequality.

Therefore, the total number of pairs (x,y) that satisfy the inequality is infinite, since there are infinitely many real numbers between -2√6 and 2√6, and each of these corresponds to a range of x values that satisfies the inequality.

Answer 2

x

Explanation:

To solve this inequality, we can use the properties of logarithms to simplify it. First, we can combine the two logarithms on the left side of the inequality using the product rule:

log₃[(x² + y² + x)(x² + y²)] ≤ log₃(x) + log₂(x² + y² + 24x)

Next, we can use the fact that logₐ(b) ≤ logₐ© if b ≤ c to simplify the right side of the inequality:

log₃[(x² + y² + x)(x² + y²)] ≤ log₃(x(x² + y² + 24x))

Now we can expand both sides of the inequality and simplify:

log₃(x⁴ + 2x³y² + x²y⁴ + x³ + xy²) ≤ log₃(x⁴ + 24x³)

Subtracting log₃(x⁴ + 24x³) from both sides gives:

log₃(x⁴ + 2x³y² + x²y⁴ + x³ + xy²) - log₃(x⁴ + 24x³) ≤ 0

Using the quotient rule for logarithms gives:

log₃[(x⁴ + 2x³y² + x²y⁴ + x³ + xy²)/(x⁴ + 24x³)] ≤ 0

Finally, we can use the fact that logₐ(b) ≤ 0 if and only if b ≤ 1 to solve for x and y:

(x⁴ + 2x³y² + x²y⁴ + x³ + xy²)/(x⁴ + 24x³) ≤ 1

Multiplying both sides by (x⁴ + 24x³) gives:

x⁴ + 2x³y² + x²y⁴ + x³ + xy² ≤ x⁴ + 24x³

Simplifying gives:

2x³y² + x²y⁴ + x³ - 24x³ ≤ -xy²

Rearranging terms gives:

xy² - 2x³y² - x²y⁴ - x³ + 24x³ ≥ 0

Factoring out an xy term gives:

xy(y - (2x)^(3/2))(y + (2x)^(3/2)) ≥ 0

This inequality holds when either y ≥ (2x)^(3/2) or y ≤ -(2x)^(3/2). Therefore, there are two pairs of numbers that satisfy this inequality for any given value of x.

I hope this helps!