Solve: $8(2^{x-3}) leq 20 - 4^x$

leq 20 - 4^x$

In the realm of mathematics, we often encounter inequalities that challenge our problem-solving skills. One such intriguing inequality is $8\left(2^{x-3}\right) leq 20-4^x$. This problem requires a deep understanding of exponential functions and how to manipulate them to find the range of values for x that satisfy the given condition. Let's embark on a journey to unravel this mathematical puzzle, exploring each step with clarity and precision. Our goal is to not only find the solution set but also to appreciate the elegance of the mathematical principles at play. We'll break down the inequality, transform it into a more manageable form, and then solve it using techniques relevant to exponential and quadratic equations. This exploration will be a testament to the power of algebraic manipulation and the systematic approach to solving complex mathematical problems. Remember, every step we take is designed to bring us closer to the ultimate solution, ensuring that we leave no stone unturned in our quest for understanding. We will also touch upon why other options might be incorrect, reinforcing our understanding of the solution.

Understanding the Inequality and Initial Transformations

The inequality we need to solve is $8\left(2^{x-3}\right) leq 20-4^x$. The first step in tackling any exponential inequality is to simplify the terms and express them with a common base. Notice that $4^x$ can be rewritten as $\left(2^2\right)^x = 2^{2x} = \left(2^x\right)^2$. This is a crucial observation because it allows us to treat the inequality as a quadratic in terms of $2^x$. Similarly, the term $8\left(2^{x-3}\right)$ can be simplified. Using the properties of exponents, $2^{x-3} = 2^x \cdot 2^{-3} = \frac{2^x}{2^3} = \frac{2^x}{8}$. Therefore, $8\left(2^{x-3}\right) = 8 \cdot \frac{2^x}{8} = 2^x$. So, our inequality transforms into: $2^x leq 20 - \left(2^x\right)^2$. This simplification is a significant stride towards solving the problem, as it brings the variable x into a more accessible form.

Substitution and Quadratic Inequality

To make the inequality even easier to work with, let's introduce a substitution. Let $y = 2^x$. Since $2^x$ is always positive for any real value of x, we know that $y > 0$. Substituting y into our transformed inequality, we get: $y leq 20 - y^2$. Now, we rearrange this inequality to bring all terms to one side, forming a standard quadratic inequality: $y^2 + y - 20 leq 0$. This is a much more familiar form, and we can solve it by finding the roots of the corresponding quadratic equation, $y^2 + y - 20 = 0$. Factoring this quadratic equation, we look for two numbers that multiply to -20 and add to 1. These numbers are 5 and -4. So, the factored form is $(y+5)(y-4) = 0$. The roots are $y = -5$ and $y = 4$. These roots divide the number line into three intervals: $(- infty, -5)$, $[-5, 4]$, and $(4, infty)$. We need to determine in which of these intervals the inequality $y^2 + y - 20 leq 0$ holds true. We can test a value from each interval. For $y < -5$, let's test $y = -6$: $(-6)^2 + (-6) - 20 = 36 - 6 - 20 = 10$, which is not $leq 0$. For $-5 leq y leq 4$, let's test $y = 0$: $(0)^2 + (0) - 20 = -20$, which is $leq 0$. For $y > 4$, let's test $y = 5$: $(5)^2 + (5) - 20 = 25 + 5 - 20 = 10$, which is not $leq 0$. Therefore, the solution to the quadratic inequality $y^2 + y - 20 leq 0$ is $-5 leq y leq 4$.

Back-Substitution and Final Solution

We have found that the solution for y is $-5 leq y leq 4$. However, remember that we made a substitution where $y = 2^x$. We must now substitute back to find the solution for x. So, we have $-5 leq 2^x leq 4$. Let's analyze this compound inequality. The first part, $-5 leq 2^x$, is always true for all real values of x because an exponential function $2^x$ is always positive. Thus, this part of the inequality does not impose any restrictions on x. The second part of the inequality is $2^x leq 4$. To solve this, we can express 4 as a power of 2: $4 = 2^2$. So, the inequality becomes $2^x leq 2^2$. Since the base of the exponential function (2) is greater than 1, the inequality holds true when the exponents satisfy the same inequality: $x leq 2$. Combining this with the fact that the first part of the inequality is always true, the solution set for x is all values less than or equal to 2. In interval notation, this is $(- infty, 2]$. This matches option B.

Analyzing the Options

Let's briefly consider why the other options are not correct.

• A. $(- infty, infty)$: This would imply that the inequality holds for all real numbers, which is not true. For example, if we take a large value of x, say $x=3$, then $8(2^{3-3}) = 8(2^0) = 8(1) = 8$. And $20 - 4^3 = 20 - 64 = -44$. Clearly, $8 leq -44$ is false.

• C. $(0, 20)$: This option represents an interval for x that is neither directly related to the exponential terms nor the solution we derived. The bounds 0 and 20 do not arise naturally from the manipulation of the inequality.

• D. $[2, infty)$: This is the opposite of our solution. If we test a value greater than 2, say $x=3$, we already saw it fails. If we test $x=2$, we get $8(2^{2-3}) = 8(2^{-1}) = 8(1/2) = 4$. And $20 - 4^2 = 20 - 16 = 4$. Here, $4 leq 4$ is true, which confirms that $x=2$ is part of the solution. However, values greater than 2 do not satisfy the inequality.

Conclusion

By carefully transforming the given exponential inequality into a quadratic inequality through substitution, and then back-substituting to find the solution for x, we have determined that the solution set is $(- infty, 2]$. This systematic approach ensures accuracy and a clear understanding of the underlying mathematical principles. The journey from a complex exponential inequality to a simple linear bound on x showcases the power and beauty of algebraic manipulation.

For further exploration into solving inequalities and understanding exponential functions, you can refer to resources like Khan Academy's mathematics sections, which offer comprehensive explanations and practice problems on these topics.