Log² ₓ₊₂(x-18)²+32 <= 16logₓ₊₂(36+16x-x²)

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вопрос

To solve this inequality, we need to first simplify both sides.

Starting with the left side of the inequality:

Log₂ ₓ₊₂(x-18)² + 32

Using the property of logarithms that logₐ(b) = c is equivalent to a^c = b, we can rewrite the left side as:

2^((x+2)-18)^2 + 32

Now, simplifying further:

2^(x-16)^2 + 32

Now, let's simplify the right side of the inequality:

16logₓ₊₂(36+16x-x²)

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

x₊₂^(36+16x-x²) = 16

Now, we have the updated inequality:

2^(x-16)^2 + 32 ≤ x₊₂^(36+16x-x²) = 16

We can continue simplifying and solving from here, depending on the specific values and relationships of x in the inequality.