Solving For Ln Y A Detailed Explanation

In this article, we will delve into a fascinating mathematical problem involving limits, square roots, and logarithms. Our primary goal is to determine the value of $\ln y$ given the limit expression: $\ln y = \lim_{x \rightarrow 9} \frac{\sqrt{x} - 3}{x - 9}$. This problem elegantly combines concepts from calculus and algebra, requiring a careful application of limit evaluation techniques. Understanding limits is crucial in calculus as they form the foundation for derivatives and integrals. The expression involves a rational function with a square root, which often necessitates algebraic manipulation to simplify and evaluate the limit. We will explore the steps required to solve this problem in detail, offering insights into the underlying principles and techniques. The problem-solving process will not only enhance your understanding of limits but also improve your algebraic skills. This exploration is essential for students and enthusiasts alike who wish to deepen their understanding of calculus and mathematical problem-solving. By the end of this article, you will have a clear understanding of how to evaluate this limit and determine the value of $\ln y$.

Understanding the Limit

Limits are a fundamental concept in calculus, representing the value that a function approaches as the input approaches a certain value. In our case, we are interested in the limit of the function $\frac{\sqrt{x} - 3}{x - 9}$ as $x$ approaches 9. Directly substituting $x = 9$ into the expression results in the indeterminate form $\frac{0}{0}$, which means we need to employ a different strategy to find the limit. Indeterminate forms like $\frac{0}{0}$ and $\frac{\infty}{\infty}$ do not provide immediate answers, and they signal the need for algebraic manipulation or the application of techniques such as L'Hôpital's Rule. The presence of the square root in the numerator suggests that rationalization might be a useful approach. Rationalization involves multiplying the numerator and denominator by the conjugate of the expression involving the square root, which helps eliminate the square root and simplify the expression. This technique is commonly used in limit problems to transform the expression into a more manageable form. By carefully manipulating the expression, we aim to eliminate the indeterminate form and reveal the true limit value. This process highlights the importance of algebraic skills in calculus, as the ability to simplify expressions is crucial for evaluating limits and solving more complex problems.

Algebraic Manipulation

To evaluate the limit $\lim_{x \rightarrow 9} \frac{\sqrt{x} - 3}{x - 9}$, the initial direct substitution leads to an indeterminate form. To resolve this, we employ algebraic manipulation, specifically rationalizing the numerator. We multiply both the numerator and the denominator by the conjugate of the numerator, which is $\sqrt{x} + 3$. This step is crucial because it eliminates the square root in the numerator, making the expression simpler to work with. The conjugate multiplication transforms the expression as follows:

$$\frac{\sqrt{x} - 3}{x - 9} \cdot \frac{\sqrt{x} + 3}{\sqrt{x} + 3} = \frac{(\sqrt{x} - 3)(\sqrt{x} + 3)}{(x - 9)(\sqrt{x} + 3)}$$

Expanding the numerator using the difference of squares formula, $(a - b)(a + b) = a^2 - b^2$, we get:

$$\frac{x - 9}{(x - 9)(\sqrt{x} + 3)}$$

Now, we can cancel the $(x - 9)$ terms from the numerator and the denominator, provided that $x \neq 9$. This simplification is valid because we are considering the limit as $x$ approaches 9, not the value of the function at $x = 9$. After canceling the terms, the expression simplifies to:

$$\frac{1}{\sqrt{x} + 3}$$

This simplified form is much easier to evaluate as $x$ approaches 9. The algebraic manipulation has effectively removed the indeterminate form, allowing us to proceed with direct substitution to find the limit.

Evaluating the Limit

After performing the algebraic manipulation, our expression has been simplified to $\frac{1}{\sqrt{x} + 3}$. Now, we can proceed to evaluate the limit as $x$ approaches 9. The simplification has made it possible to use direct substitution, which is a straightforward method for evaluating limits when the function is continuous at the point of interest. Substituting $x = 9$ into the simplified expression, we get:

$$\lim_{x \rightarrow 9} \frac{1}{\sqrt{x} + 3} = \frac{1}{\sqrt{9} + 3}$$

Since $\sqrt{9} = 3$, the expression further simplifies to:

$$\frac{1}{3 + 3} = \frac{1}{6}$$

Therefore, the limit of the original expression as $x$ approaches 9 is $\frac{1}{6}$. This means that $\ln y = \frac{1}{6}$. Evaluating the limit in this way demonstrates the power of algebraic manipulation in simplifying expressions and making them amenable to direct substitution. The ability to recognize and apply appropriate algebraic techniques is a key skill in calculus and is essential for solving a wide range of limit problems.

Solving for y

Having determined that $\ln y = \frac{1}{6}$, our next step is to solve for $y$. The natural logarithm, denoted as $\ln$, is the logarithm to the base $e$, where $e$ is the mathematical constant approximately equal to 2.71828. To find $y$, we need to use the inverse operation of the natural logarithm, which is the exponential function. The relationship between logarithms and exponentials is fundamental in mathematics and is crucial for solving equations involving logarithms. To isolate $y$, we raise $e$ to the power of both sides of the equation $\ln y = \frac{1}{6}$. This operation effectively undoes the natural logarithm on the left side, leaving us with $y$ on its own. The equation becomes:

$$e^{\ln y} = e^{\frac{1}{6}}$$

By the definition of logarithms, $e^{\ln y} = y$, so we have:

$$y = e^{\frac{1}{6}}$$

This expression gives us the exact value of $y$. The value $e^{\frac{1}{6}}$ represents the sixth root of $e$. It's important to recognize that this is the precise solution, and while we could approximate it using a calculator, leaving the answer in this form preserves its exactness. Understanding how to convert between logarithmic and exponential forms is a vital skill in mathematics, particularly in calculus and algebra. This step demonstrates how we can use this relationship to solve for variables within logarithmic equations.

Final Answer

In summary, we have successfully determined the value of $\ln y$ and subsequently solved for $y$. Starting with the limit expression $\ln y = \lim_{x \rightarrow 9} \frac{\sqrt{x} - 3}{x - 9}$, we encountered an indeterminate form upon direct substitution. To overcome this, we employed the technique of rationalizing the numerator, which involved multiplying both the numerator and the denominator by the conjugate of the numerator. This algebraic manipulation allowed us to simplify the expression and eliminate the indeterminate form. After simplification, we evaluated the limit by direct substitution, finding that $\lim_{x \rightarrow 9} \frac{\sqrt{x} - 3}{x - 9} = \frac{1}{6}$. Therefore, we concluded that $\ln y = \frac{1}{6}$. To find $y$, we used the inverse relationship between logarithms and exponentials, raising $e$ to the power of both sides of the equation. This led us to the solution $y = e^{\frac{1}{6}}$. This final answer represents the exact value of $y$, expressing it in terms of the mathematical constant $e$. This exercise has demonstrated the importance of understanding limits, algebraic manipulation, and the relationship between logarithmic and exponential functions in solving mathematical problems. The systematic approach we followed highlights the key steps involved in evaluating limits and solving equations involving logarithms and exponentials. This detailed solution provides a comprehensive understanding of the problem and its solution, which is beneficial for students and anyone interested in mathematical problem-solving.

In conclusion, we have successfully navigated through the problem of finding $\ln y$ given the limit expression $\ln y = \lim_{x \rightarrow 9} \frac{\sqrt{x} - 3}{x - 9}$. Our journey involved several key steps, each building upon the previous one. We began by recognizing the indeterminate form that arises from direct substitution, which necessitated the use of algebraic manipulation. The technique of rationalizing the numerator proved to be crucial in simplifying the expression, allowing us to eliminate the indeterminate form. Following this, we evaluated the limit by direct substitution, a method made possible by our earlier simplification efforts. This yielded the value of $\ln y$, which we then used to solve for $y$ by employing the inverse relationship between logarithms and exponentials. The final solution, $y = e^{\frac{1}{6}}$, elegantly expresses $y$ in terms of the mathematical constant $e$. This problem serves as an excellent illustration of the interconnectedness of various mathematical concepts. It highlights the importance of a solid foundation in algebra, particularly in manipulating expressions and recognizing patterns. It also underscores the significance of understanding limits, a cornerstone of calculus, and the relationship between logarithmic and exponential functions. The systematic approach we adopted in solving this problem can be applied to a wide range of mathematical challenges. By breaking down the problem into smaller, manageable steps, we were able to address each aspect methodically and arrive at the solution. This process emphasizes the value of careful analysis, strategic problem-solving, and a thorough understanding of fundamental mathematical principles. This exercise has not only provided a solution to a specific problem but has also reinforced broader mathematical skills and concepts, which are essential for further study and exploration in mathematics and related fields.