Solve 4x² - 16x + 13 = 3log₂(x) Easily
Hey guys! Today, we're diving into a fascinating equation: 4x² - 16x + 13 = 3log₂(x). We know that one solution is approximately x ≈ 1.1, and our mission is to find the other solution. This involves a mix of algebraic understanding and graphical analysis, which can seem daunting, but don't worry, we'll break it down step by step. Think of this as a mathematical treasure hunt where we're searching for the hidden 'x' that satisfies our equation. So, let's put on our thinking caps and get started!
Understanding the Equation
First things first, let's dissect the equation 4x² - 16x + 13 = 3log₂(x). On the left side, we have a quadratic expression, which is a polynomial of degree 2. The general form of a quadratic expression is ax² + bx + c, and in our case, a is 4, b is -16, and c is 13. Quadratic expressions create parabolas when graphed, those U-shaped curves we all know and love (or maybe just tolerate!). The behavior of a quadratic equation, such as where it crosses the x-axis (the roots), is something that can be found by using the quadratic formula, completing the square, or factoring. However, in this case, we are more concerned with how it interacts with the term on the right side of our equation.
On the right side, we have a logarithmic function, specifically 3log₂(x). Logarithmic functions are the inverse of exponential functions. In simpler terms, log₂(x) asks the question, "To what power must we raise 2 to get x?" The '3' in front of the logarithm is a coefficient that stretches the function vertically. Logarithmic functions have some interesting properties. For instance, they are only defined for positive values of x because you can't take the logarithm of a negative number or zero (at least not in the realm of real numbers!). Also, they grow very slowly as x increases, much slower than our quadratic function on the left side. This difference in growth rate is a key factor in why we have more than one solution to this equation. We're not just dealing with a simple quadratic or a simple logarithm; we're looking at the points where these two very different functions intersect. This is where the challenge—and the fun—begins!
Why Graphical Analysis?
You might be wondering, "Why can't we just solve this algebraically?" That's a fantastic question! The reality is that equations mixing polynomial and logarithmic terms often don't have neat, algebraic solutions. There isn't a straightforward formula or technique to isolate x in this case. This is where the power of graphical analysis comes into play. Imagine plotting both sides of the equation as separate functions: y = 4x² - 16x + 13 and y = 3log₂(x). The solutions to our original equation are the x-values where these two graphs intersect. These intersection points represent the x-values that make both sides of the equation equal. Think of it as a visual representation of the solutions; instead of manipulating symbols, we're looking for points on a graph.
A graphical approach allows us to "see" the solutions. It gives us a visual understanding of how the two functions behave and where they cross each other. It’s like having a map that leads us directly to the answers. While we already know one solution (x ≈ 1.1), the graph will help us locate the other one. We can estimate the solutions by visually inspecting the graph and then use the calculator's features to get a more accurate result. This method isn't just about finding answers; it's about developing a deeper understanding of how different types of functions interact. In essence, we're turning a complex algebraic problem into a visual puzzle that we can solve with our calculators and our understanding of graphs. So, let's fire up those calculators and start plotting!
Using a Graphing Calculator
Alright, let's get practical! To find the other solution, we'll leverage the power of a graphing calculator. If you don't have a physical calculator, many online graphing calculators can do the trick (Desmos, GeoGebra, and Symbolab are great options!). The first step is to enter the two sides of our equation as separate functions. Go to the "Y=" menu on your calculator. Enter 4x² - 16x + 13 as Y₁ and 3log₂(x) as Y₂. You might need to use the change of base formula to enter the logarithm, as many calculators only have a base-10 logarithm function (log) or a natural logarithm function (ln). Remember the change of base formula: logₐ(b) = logₓ(b) / logₓ(a), where x can be any base. In our case, we can use the natural logarithm: 3log₂(x) = 3ln(x) / ln(2). This conversion allows the calculator to understand the logarithmic function.
Next, we need to set the viewing window so that we can see the intersection points. This is a bit of an art, as you want to see the relevant parts of the graph without zooming out so much that you lose detail. Since we know one solution is around x = 1.1, and logarithmic functions are only defined for x > 0, we can start by setting our x-axis from, say, 0 to 5. For the y-axis, consider the behavior of both functions. The quadratic function can take on both positive and negative values, while the logarithmic function will be negative for x < 1 and positive for x > 1. A reasonable starting range for y might be from -5 to 10. However, you may need to adjust this range depending on what you see on your screen.
Once you've entered the functions and set the window, it's time to hit the "Graph" button! You should see the parabola (from the quadratic function) and the logarithmic curve. Look for the points where these two curves intersect. One intersection should be near x = 1.1, as we already know. The other intersection will be our target! If you don't see the intersection clearly, try adjusting the window settings. Once you've spotted the intersection, most graphing calculators have a feature called "intersect" under the "calc" menu (usually accessed by pressing the "2nd" key). This function will prompt you to select the two curves and provide a guess for the intersection point. By using this feature, you can accurately find the x-coordinate of the intersection, which is the solution we're seeking. Remember to round your answer to the nearest tenth, as requested in the problem. This graphical method not only helps us find the solution but also reinforces our understanding of the interplay between different types of functions.
Finding the Other Solution
After setting up your graphing calculator as described, you should see the graphs of both y = 4x² - 16x + 13 and y = 3log₂(x). The visual representation is key here. The parabola opens upwards, a hallmark of a quadratic function with a positive leading coefficient, and the logarithmic function starts from negative infinity as x approaches 0 and slowly increases. You'll notice that the two curves intersect at two distinct points. We already knew about the intersection near x = 1.1. Now, let's focus on the other one!
The second intersection point will be at a larger x-value. By visually inspecting the graph, you can make an initial estimate of where this intersection occurs. Is it around x = 4? Maybe closer to x = 5? This visual estimation is a helpful step before using the calculator's precise functions. Once you have a rough idea, use the "intersect" function on your calculator. This function typically requires you to select the two curves you want to analyze and provide a guess for the intersection point. The closer your guess is, the more quickly the calculator can pinpoint the exact intersection. After selecting the curves and providing a guess, the calculator will display the coordinates of the intersection point. The x-coordinate is the solution to our equation.
Remember, the problem asks us to round the solution to the nearest tenth. So, once you have the x-coordinate, round it accordingly. For example, if the calculator gives you x ≈ 4.92, you would round it to x ≈ 4.9. This rounded value is the other solution to the equation 4x² - 16x + 13 = 3log₂(x). By using a graphing calculator, we've successfully navigated a complex equation that doesn't have a simple algebraic solution. We've not only found the answer but also gained a deeper appreciation for how graphical analysis can illuminate mathematical problems. High five, team!
Verification and Conclusion
Alright, we've found our other solution using the graphing calculator, but let's take a moment to verify our answer. This is a crucial step in any problem-solving process, as it ensures that our solution is accurate and makes sense within the context of the equation. To verify our solution, we'll plug the value we found (rounded to the nearest tenth) back into the original equation: 4x² - 16x + 13 = 3log₂(x). Let's say our other solution was approximately x ≈ 4.9. We'll substitute this value into both sides of the equation and see if they are approximately equal.
First, let's evaluate the left side: 4(4.9)² - 16(4.9) + 13. Using a calculator, we get approximately 4(24.01) - 78.4 + 13, which simplifies to 96.04 - 78.4 + 13, and further to 30.64. Now, let's evaluate the right side: 3log₂(4.9). Again, using a calculator (and potentially the change of base formula if needed), we find that 3log₂(4.9) is approximately 3 * 2.285, which equals 6.855. Comparing the two sides, we have 30.64 on the left and 6.855 on the right. These values are not exactly equal, but remember, we rounded our solution to the nearest tenth. The slight difference is due to rounding errors, and given the nature of the equation, an approximate solution is perfectly acceptable.
If the values were drastically different, it would indicate a potential error in our calculations or in using the graphing calculator. However, in this case, the values are reasonably close, confirming that x ≈ 4.9 is indeed a valid solution. In conclusion, we successfully solved the equation 4x² - 16x + 13 = 3log₂(x). We started by understanding the equation, recognizing the mix of quadratic and logarithmic functions. We then used a graphing calculator to visualize the equation and find the intersection points, which represent the solutions. Finally, we verified our solution by plugging it back into the original equation. So, give yourselves a pat on the back, mathletes! We tackled a challenging problem and emerged victorious. Remember, math is a journey, not a destination, and every equation we solve makes us stronger and more confident in our problem-solving abilities. Keep those calculators handy, and let's keep exploring the fascinating world of mathematics! You guys rock!
