Question

Log2 x + logx 2 ≤ 2.5

Answer 1

Ответ:там всьо по порядку на фотках самая последняя это первое записывай

Объяснение:

Answer 2

To solve the inequality logarithmically, we can use the properties of logarithms. The given inequality is:

log2(x) + logx(2) ≤ 2.5

To simplify this expression, we can use the logarithmic identities. Recall that log base a of b is equal to log base c of b divided by log base c of a. Using this property, we can rewrite the inequality as:

log(x) / log(2) + log(2) / log(x) ≤ 2.5

Now, let's simplify further. We can multiply the entire inequality by log(2) to eliminate the denominators:

log(x) + (log(2))^2 / log(x) ≤ 2.5 * log(2)

log(x) + log(2)^2 / log(x) ≤ 2.5 * log(2)

log(x) + 4 / log(x) ≤ 2.5 * log(2)

Now, let's combine the terms on the left-hand side:

(log(x)^2 + 4) / log(x) ≤ 2.5 * log(2)

Next, we can multiply both sides of the inequality by log(x) to eliminate the denominator:

log(x)^2 + 4 ≤ 2.5 * log(2) * log(x)

Now, rearrange the terms:

log(x)^2 - 2.5 * log(2) * log(x) + 4 ≤ 0

This equation is a quadratic equation in terms of log(x). We can solve this quadratic equation by substituting log(x) with a variable, let's say u. The equation becomes:

u^2 - 2.5 * log(2) * u + 4 ≤ 0

Now, solve this quadratic equation for u using the quadratic formula:

u = (-b ± sqrt(b^2 - 4ac)) / 2a

where a = 1, b = -2.5 * log(2), and c = 4. Substitute the values:

u = (-(-2.5 * log(2)) ± sqrt((-2.5 * log(2))^2 - 4 * 1 * 4)) / 2 * 1

Simplifying further:

u = (2.5 * log(2) ± sqrt((2.5 * log(2))^2 - 16)) / 2

Now that we have the value of u, we need to substitute it back to log(x) in order to find the solutions for x. However, it seems there's an error in the equation you provided. The terms log(2) and log(x) cannot be directly added together because they have different bases. Could you please double-check the equation or provide more context?