Solving the Equation 4x 4 × 2^x: A Comprehensive Guide

When dealing with equations like 4x 4 × 2^x, it is crucial to not only find the solution, but also to determine the number of possible solutions and the intervals in which they might lie. In this article, we will explore these aspects in detail using both graphical and analytical methods, concluding with a numerical approach to find the exact solution.

Graphical Insight

Step 1: Plotting the Functions
To get a qualitative understanding of the solutions, we first plot the left-hand side (LHS) and right-hand side (RHS) of the equation on the same set of axes. The LHS is a straight line given by 4x, while the RHS is an exponential function 4 × 2^x. By observing the intersections of these two functions, we can estimate the number and approximate values of the solutions.

The exponential function 4 × 2^x is always positive and increasing with x, representing a curve that curves upward. This curve will intersect the straight line 4x at certain points, depending on the value of x. For instance, if the line is shifted up or down, the number of intersections may change, giving us a range of possible solutions.

Analytical Insight

Step 2: Defining the Function
To determine the exact number of solutions, let's define a function f(x) that equates the LHS and RHS of the original equation:

[ f(x) 4 times 2^x - 4x - 4 ]

Solving f(x) 0 would give the exact solutions to the original equation. The next step is to find the derivative of f(x) to understand its critical points:

[ f'(x) 4 times 2^x ln(2) - 1 ]

Step 3: Analyzing the Derivative
The derivative f'(x) shows that the function has exactly one critical point where it changes from decreasing to increasing. This critical point occurs at:

[ x^* -frac{ln 4}{ln 2} approx -1.02 ]

At this point, the function has a local minimum. To determine the exact number of solutions, we need to evaluate the function at this point and at the boundaries of the intervals around it.

Numerical Insight

Step 4: Evaluating the Function
First, let's evaluate the function at the critical point:

[ f(-1) 4 times 2^{-1} - 4(-1) - 4 2 4 - 4 2 ]

Next, we need to check the function values on either side of the critical point to determine the behavior:

[ f(-2) 4 times 2^{-2} - 4(-2) - 4 1 8 - 4 5 ]

[ f(-3) 4 times 2^{-3} - 4(-3) - 4 0.5 12 - 4 8.5 ]

Since all these values are positive, and considering the derivative, we can conclude that there is no change in sign before the critical point, indicating a single solution at x 0.

Further, we can try values close to -1.02 to find the second solution:

[ f(-4) 4 times 2^{-4} - 4(-4) - 4 0.125 16 - 4 12.125 ]

[ f(-3.5) 4 times 2^{-3.5} - 4(-3.5) - 4 approx 0.353 14 - 4 10.353 ]

[ f(-3) 8.5 ]

[ f(-2.5) 4 times 2^{-2.5} - 4(-2.5) - 4 approx 0.577 10 - 4 6.577 ]

[ f(-2) 5 ]

[ f(-1.5) 4 times 2^{-1.5} - 4(-1.5) - 4 approx 1.732 6 - 4 3.732 ]

[ f(-1) 2 ]

By checking these values, we find that a sign change occurs between f(-2) and f(-1.5), indicating that there is a second solution in the interval (-2, -1.5).

Conclusion: Therefore, the solutions are x 0 and a second solution in the interval (-2, -1.5). This solution can be refined further using numerical methods, such as Newton's method, to find a more precise value.