How to Solve This Inequality and Related Concepts

Solving inequalities is an essential skill in mathematics, especially when dealing with logarithmic and exponential functions. This article will guide you through the process of solving the inequality given, and we will explore the underlying concepts of logarithmic and exponential functions.

Understanding the Given Inequality

Consider the inequality:

(frac{1}{16} leq 2^{log_2 x - log_2 x - log_2 x})

This can be simplified by taking the logarithm base two of both sides, which gives us an equivalent inequality.

Solving the Inequality

To solve the inequality, let's start by taking the logarithm base two of both sides:

(log_2left(frac{1}{16}right) leq log_2(2^{log_2 x - log_2 x - log_2 x}))

Recall that (log_2(2^{log_2 x}) log_2 x). Therefore, the right-hand side simplifies to:

(log_2(2^{log_2 x - log_2 x - log_2 x}) log_2 x - log_2 x - log_2 x)

Using the properties of logarithms, rewrite (log_2 x - log_2 x - log_2 x) as:

(-log_2 x^3)

Substituting and Simplifying

Set (log_2 x y). Then the inequality becomes:

(log_2(frac{1}{16}) leq -y^3)

Simplify the left-hand side:

(log_2(frac{1}{16}) -4)

So the inequality is:

(-4 leq -y^3)

Multiply both sides by -1 (and reverse the inequality):

(4 geq y^3)

Solving the Cube Inequality

The inequality (4 geq y^3) can be rewritten as a cubic equation:

(y^3 - 4 leq 0)

Factor the left-hand side:

((y - 1)(y 2)^2 leq 0)

This inequality is satisfied for:

(y leq 1) or (y geq -2)

Since (y log_2 x), the solutions are:

(log_2 x leq 1) or (log_2 x geq -2)

Exponentiating both sides with base 2, we get:

(x leq 2^1) or (x geq 2^{-2})

Therefore, the solution is:

(x leq 2) or (x geq frac{1}{4})

Expressing this in interval notation:

((0, frac{1}{4}] U [2, infty))

Revisiting the Solution with Another Approach

We can also solve the inequality by setting (log_2 x z). Then the inequality becomes:

(2^z leq 2^{z - z - z})

Simplify the right-hand side:

(2^z leq 2^{-z})

Take the logarithm base two of both sides:

(z leq -z)

simplify to:

(2z leq 0)

Divide by 2:

(z leq 0)

Express (z log_2 x):

(log_2 x leq 0)

Exponentiating both sides with base 2:

(x leq 2^0 1)

This gives us the interval:

(0 leq x leq 1)

However, the original problem was (frac{1}{16} leq 2^{log_2 x - log_2 x - log_2 x}), which simplifies to:

(frac{1}{16} leq 2^{-3log_2 x})

This simplifies to:

(2^{-4} leq 2^{-3log_2 x})

Solving for (x), we get:

(4 geq 3log_2 x)

Divide by 3:

(log_2 x leq -frac{4}{3})

Exponentiate with base 2:

(x leq 2^{-frac{4}{3}})

This gives us the interval:

((0, 2^{-frac{4}{3}}])

Again, combining all intervals and intervals, the solution is:

(-infty leq x leq -frac{1}{2} text{ or } 4 leq x leq infty)

Expressing in interval notation:

((-infty, -frac{1}{2}] U [4, infty))

Conclusion

Solving the inequality (frac{1}{16} leq 2^{log_2 x - log_2 x - log_2 x}) involves understanding the properties of logarithmic and exponential functions. By breaking down the problem into smaller, manageable parts and carefully applying the rules of logarithms and exponents, we can arrive at the correct solution. This process not only solves the inequality but also reinforces key mathematical concepts.

Related Keywords

inequality solving logarithmic functions exponential functions mathematical solutions