Finding The Tangent A Value For Y = Xln3 + 2a And Y = 3^x + 7

For what value of 'a' is the line y = xln3 + 2a tangent to the graph of the function y = 3^x + 7?

Introduction

In this article, we delve into a fascinating problem involving exponential and linear functions. Our primary goal is to determine the specific value of 'a' for which the straight line y = xln3 + 2a becomes a tangent to the exponential curve y = 3^x + 7. This problem elegantly combines concepts from algebra and calculus, requiring a solid understanding of derivatives and tangent lines. To solve this, we'll explore the conditions necessary for a line to be tangent to a curve, utilizing derivatives to find the slope of the tangent and setting up equations to solve for the unknown 'a'. This exploration will not only provide a solution to the problem but also offer valuable insights into the interplay between different types of functions and their graphical representations.

Understanding the Problem

Before diving into the solution, it's crucial to fully grasp the problem statement. We are given two functions: a linear function y = xln3 + 2a and an exponential function y = 3^x + 7. The linear function represents a straight line with a slope of ln3 and a y-intercept of 2a. The exponential function represents a curve that grows rapidly as x increases. The core question is: For what value of a will the line y = xln3 + 2a touch the curve y = 3^x + 7 at exactly one point, i.e., be tangent to it?

This tangency condition implies two key things: firstly, at the point of tangency, the y-values of both functions must be equal. Secondly, the slopes of the line and the curve must also be equal at that same point. This is because the tangent line, by definition, has the same slope as the curve at the point of contact. Mathematically, this translates to the derivative of the exponential function being equal to the slope of the line at the point of tangency. By setting up these two equations, we can solve for the unknown x-coordinate of the point of tangency and, subsequently, the value of a. This approach provides a systematic way to tackle problems involving tangent lines and curves, highlighting the power of calculus in solving geometric problems.

Setting up the Equations

To solve this problem, we'll establish two crucial equations based on the conditions of tangency. Let's denote the point of tangency as (x₀, y₀). At this point, both the function values and the slopes must be equal.

1. Equality of Function Values: This condition implies that at x = x₀, the y-values of both functions must be the same. Therefore, we have:

   3^x₀ + 7 = x₀ln3 + 2a

   This equation represents the fact that the line and the curve intersect at the point of tangency. It's our first key relationship connecting x₀ and a.

2. Equality of Slopes: For the line to be tangent to the curve, their slopes must be equal at the point of tangency. The slope of the line y = xln3 + 2a is simply ln3. To find the slope of the curve y = 3^x + 7, we need to calculate its derivative:

   dy/dx = d(3^x + 7)/dx = 3^xln3

   At the point of tangency (x₀, y₀), the slope of the curve is 3^x₀ln3. Setting this equal to the slope of the line, we get:

   3^x₀ln3 = ln3

   This equation is a direct consequence of the tangency condition and provides a critical link between the exponential function and its tangent line. By solving these two equations simultaneously, we can determine the values of x₀ and a that satisfy the tangency condition. This structured approach allows us to break down a complex problem into manageable steps, leveraging the principles of calculus and algebra to arrive at a solution.

Solving for x₀

Our next step involves solving the equation derived from the equality of slopes: 3^x₀ln3 = ln3. This equation is relatively straightforward to solve and will give us the x-coordinate of the point of tangency.

To isolate 3^x₀, we can divide both sides of the equation by ln3. Since ln3 is a non-zero constant (approximately 1.0986), this operation is valid and yields:

3^x₀ = 1

Now, we need to determine the value of x₀ that satisfies this equation. We know that any non-zero number raised to the power of 0 equals 1. Therefore:

x₀ = 0

This result is significant as it tells us that the point of tangency occurs at x = 0. This means the line y = xln3 + 2a touches the curve y = 3^x + 7 when x is 0. This discovery simplifies the subsequent steps, as we can now substitute this value into the other equation to solve for a. The ability to pinpoint the x-coordinate of the point of tangency is a crucial step in solving this problem, highlighting the power of algebraic manipulation in simplifying complex equations.

Solving for a

Now that we've determined x₀ = 0, we can substitute this value back into the equation representing the equality of function values:

3^x₀ + 7 = x₀ln3 + 2a

Substituting x₀ = 0 gives us:

3⁰ + 7 = 0 * ln3 + 2a

Simplifying this equation, we know that 3⁰ = 1 and 0 * ln3 = 0. Therefore, the equation becomes:

1 + 7 = 2a

8 = 2a

To solve for a, we divide both sides of the equation by 2:

a = 4

This is our final answer. The value of a for which the line y = xln3 + 2a is tangent to the curve y = 3^x + 7 is a = 4. This solution demonstrates how we can use the principles of calculus and algebra to solve problems involving tangent lines and curves. By setting up appropriate equations based on the conditions of tangency and solving them systematically, we can arrive at the desired result. This process not only provides the answer but also reinforces our understanding of the relationships between different types of functions and their graphical representations.

Verifying the Solution

To ensure the accuracy of our solution, it's essential to verify that the line y = xln3 + 2a, with a = 4, is indeed tangent to the curve y = 3^x + 7. This verification process reinforces our understanding of the problem and the solution method.

Substituting a = 4 into the equation of the line, we get:

y = xln3 + 2(4)

y = xln3 + 8

Now, we need to check if this line is tangent to the curve y = 3^x + 7. We already found that the point of tangency occurs at x₀ = 0. Let's find the y-coordinate of this point on the curve:

y₀ = 3⁰ + 7 = 1 + 7 = 8

So, the point of tangency is (0, 8). Now, we need to check if this point also lies on the line y = xln3 + 8:

8 = 0 * ln3 + 8

8 = 8

This confirms that the point (0, 8) lies on the line. Furthermore, we already established that the slope of the line is ln3 and the slope of the curve at x = 0 is 3⁰ln3 = ln3. Since the line and the curve share a common point and have the same slope at that point, we can confidently conclude that the line y = xln3 + 8 is indeed tangent to the curve y = 3^x + 7. This verification step provides a robust check on our solution, ensuring its correctness and solidifying our understanding of the concepts involved.

Conclusion

In conclusion, we have successfully determined the value of a for which the line y = xln3 + 2a is tangent to the curve y = 3^x + 7. By applying the principles of calculus and algebra, we found that the value of a is 4. This solution was achieved by first setting up two equations based on the conditions of tangency: the equality of function values and the equality of slopes. We then solved these equations systematically, first finding the x-coordinate of the point of tangency and then using this value to solve for a.

The verification process further confirmed the accuracy of our solution, demonstrating that the line y = xln3 + 8 is indeed tangent to the curve y = 3^x + 7 at the point (0, 8). This problem exemplifies the power of mathematical tools in solving geometric problems and highlights the importance of a structured approach in tackling complex questions. The interplay between algebraic manipulation and calculus concepts, such as derivatives, is crucial in understanding and solving such problems. This exploration not only provides a specific answer but also deepens our understanding of the relationships between different types of functions and their graphical representations, fostering a more comprehensive appreciation of mathematics.