Condensing Logarithms, Proving Hyperbolic Identities, And Calculating Cylinder Volume

In the realm of mathematics, condensing logarithmic expressions is a fundamental skill, essential for simplifying complex equations and solving various problems. In this section, we'll delve into the process of condensing the expression $5 \log _3(x)+2 \log _3(4 x)-\log _3 8 x^5$, providing a step-by-step guide to mastering this technique. To effectively condense logarithmic expressions, a solid understanding of the properties of logarithms is crucial. These properties act as the building blocks for simplification and condensation. The key properties we'll utilize include:

1. The Power Rule: This rule states that $\log_b(x^n) = n \log_b(x)$, allowing us to move exponents within logarithms.
2. The Product Rule: This rule states that $\log_b(xy) = \log_b(x) + \log_b(y)$, enabling us to combine the logarithms of products.
3. The Quotient Rule: This rule states that $\log_b(\frac{x}{y}) = \log_b(x) - \log_b(y)$, allowing us to combine the logarithms of quotients.

Applying the power rule to the given expression, we begin by moving the coefficients as exponents:

$5 \log _3(x) + 2 \log _3(4 x) - \log _3(8 x^5) = \log _3(x^5) + \log _3((4x)^2) - \log _3(8x^5)$

Next, we simplify the term $(4x)^2$:

$\log _3(x^5) + \log _3((4x)^2) - \log _3(8x^5) = \log _3(x^5) + \log _3(16x^2) - \log _3(8x^5)$

Now, we employ the product rule to combine the first two logarithmic terms:

$\log _3(x^5) + \log _3(16x^2) - \log _3(8x^5) = \log _3(x^5 * 16x^2) - \log _3(8x^5)$

Simplifying the product inside the logarithm:

$\log _3(x^5 * 16x^2) - \log _3(8x^5) = \log _3(16x^7) - \log _3(8x^5)$

Finally, we apply the quotient rule to condense the expression into a single logarithm:

$\log _3(16x^7) - \log _3(8x^5) = \log _3(\frac{16x^7}{8x^5})$

Simplifying the fraction within the logarithm yields the fully condensed expression:

$\log _3(\frac{16x^7}{8x^5}) = \log _3(2x^2)$

Therefore, the condensed form of the expression $5 \log _3(x)+2 \log _3(4 x)-\log _3 8 x^5$ is $\log _3(2x^2)$. This step-by-step process demonstrates the power of logarithmic properties in simplifying complex expressions. Mastering these properties allows for efficient manipulation and solution of logarithmic equations. Understanding these rules is also fundamental for tackling more advanced mathematical concepts, making it a crucial skill for students and professionals alike. This condensation process not only simplifies the expression but also provides a clearer understanding of the relationship between the variables involved. Logarithms are a powerful tool in mathematics, and the ability to manipulate them effectively is key to success in various fields. From solving exponential equations to modeling real-world phenomena, logarithms play a significant role. By practicing and applying these properties, one can gain confidence in handling logarithmic expressions and unlock their full potential. The process of condensing logarithmic expressions is not just a mathematical exercise; it's a skill that enhances problem-solving abilities and deepens the understanding of mathematical relationships. By breaking down complex expressions into simpler forms, we can gain insights that might otherwise be obscured. This is the essence of mathematical simplification, and it's a skill that is applicable across various domains. Keep practicing, and you'll find that condensing logarithmic expressions becomes second nature, opening doors to more advanced mathematical concepts and applications.

Hyperbolic functions, while less commonly encountered than their trigonometric counterparts, play a significant role in various areas of mathematics, physics, and engineering. In this section, we will embark on a journey to prove the fundamental hyperbolic identity $ch^2 - sh^2 = 1$. This identity, analogous to the trigonometric identity $\cos^2(x) + \sin^2(x) = 1$, forms the cornerstone of many hyperbolic function relationships. To prove this identity, we must first understand the definitions of the hyperbolic functions involved. The hyperbolic cosine, denoted as $ch(x)$, and the hyperbolic sine, denoted as $sh(x)$, are defined as follows:

• $ch(x) = \frac{e^x + e^{-x}}{2}$
• $sh(x) = \frac{e^x - e^{-x}}{2}$

These definitions, based on the exponential function, provide the foundation for exploring the properties and identities of hyperbolic functions. With these definitions in hand, we can proceed to prove the identity $ch^2 - sh^2 = 1$. The proof involves substituting the exponential definitions of $ch(x)$ and $sh(x)$ into the equation and simplifying the resulting expression.

Starting with the left-hand side of the equation, $ch^2(x) - sh^2(x)$, we substitute the definitions:

$ch^2(x) - sh^2(x) = (\frac{e^x + e^{-x}}{2})^2 - (\frac{e^x - e^{-x}}{2})^2$

Next, we expand the squares:

$(\frac{e^x + e^{-x}}{2})^2 - (\frac{e^x - e^{-x}}{2})^2 = \frac{(e^x + e^{-x})^2}{4} - \frac{(e^x - e^{-x})^2}{4}$

$\frac{(e^x + e^{-x})^2}{4} - \frac{(e^x - e^{-x})^2}{4} = \frac{e^{2x} + 2e^xe^{-x} + e^{-2x}}{4} - \frac{e^{2x} - 2e^xe^{-x} + e^{-2x}}{4}$

Recall that $e^x * e^{-x} = e^{x-x} = e^0 = 1$. Substituting this into the expression:

$\frac{e^{2x} + 2e^xe^{-x} + e^{-2x}}{4} - \frac{e^{2x} - 2e^xe^{-x} + e^{-2x}}{4} = \frac{e^{2x} + 2 + e^{-2x}}{4} - \frac{e^{2x} - 2 + e^{-2x}}{4}$

Now, we combine the fractions:

$\frac{e^{2x} + 2 + e^{-2x}}{4} - \frac{e^{2x} - 2 + e^{-2x}}{4} = \frac{(e^{2x} + 2 + e^{-2x}) - (e^{2x} - 2 + e^{-2x})}{4}$

Simplifying the numerator by distributing the negative sign:

$\frac{(e^{2x} + 2 + e^{-2x}) - (e^{2x} - 2 + e^{-2x})}{4} = \frac{e^{2x} + 2 + e^{-2x} - e^{2x} + 2 - e^{-2x}}{4}$

Notice that $e^{2x}$ and $-e^{2x}$ cancel out, and $e^{-2x}$ and $-e^{-2x}$ also cancel out. We are left with:

$\frac{e^{2x} + 2 + e^{-2x} - e^{2x} + 2 - e^{-2x}}{4} = \frac{4}{4}$

Finally, simplifying the fraction:

$\frac{4}{4} = 1$

Thus, we have shown that $ch^2(x) - sh^2(x) = 1$, which is the right-hand side of the identity. This completes the proof. This identity, and others like it, are fundamental in understanding the behavior and applications of hyperbolic functions. Hyperbolic functions are not just abstract mathematical concepts; they have practical applications in various fields. For instance, they are used in physics to describe the shape of a hanging cable (catenary curve) and in engineering to model certain types of waves. Understanding the identities and relationships between hyperbolic functions is crucial for these applications. The proof we've just undertaken highlights the power of definitions in mathematics. By starting with the fundamental definitions of the hyperbolic functions, we were able to manipulate the expressions and arrive at the desired result. This approach is common in mathematical proofs and demonstrates the importance of a solid understanding of basic concepts. Furthermore, this proof showcases the elegance and interconnectedness of mathematical ideas. The relationship between exponential functions and hyperbolic functions is a testament to the underlying unity of mathematics. By exploring these connections, we gain a deeper appreciation for the subject and its applications. Keep exploring, keep proving, and you'll discover the beauty and power of mathematics. The journey of mathematical discovery is a rewarding one, filled with challenges and triumphs. Each proof completed, each identity understood, brings us closer to a deeper understanding of the world around us. So, embrace the challenge, and continue to explore the fascinating world of mathematics. The ability to prove identities is a critical skill in mathematics. It requires a combination of knowledge, logical reasoning, and algebraic manipulation. By mastering these skills, you'll be well-equipped to tackle more advanced mathematical problems and gain a deeper understanding of the subject.

In the realm of geometry, calculating the volume of three-dimensional shapes is a fundamental skill with numerous practical applications. In this section, we'll focus on calculating the volume of a cylinder, a shape commonly encountered in everyday life, from storage tanks to beverage cans. We'll specifically address the scenario of a cylindrical container with a radius of 7 meters and a vertical height of 20 meters, calculating the amount of water it can hold. The volume of a cylinder is determined by the formula:

$V = \pi r^2 h$

where:

• $V$ represents the volume of the cylinder.
• $r$ represents the radius of the circular base.
• $h$ represents the height of the cylinder.
• $\pi$ (pi) is a mathematical constant approximately equal to 3.14159.

This formula stems from the basic principle that the volume of a prism (which includes cylinders) is the area of the base multiplied by the height. In the case of a cylinder, the base is a circle, and its area is given by $\pi r^2$. Therefore, multiplying this area by the height $h$ gives us the volume of the cylinder. Given the dimensions of the cylindrical container, we have a radius $r = 7$ meters and a height $h = 20$ meters. To calculate the volume, we substitute these values into the formula:

$V = \pi (7 \text{ m})^2 (20 \text{ m})$

First, we square the radius:

$V = \pi (49 \text{ m}^2) (20 \text{ m})$

Next, we multiply the squared radius by the height:

$V = \pi (980 \text{ m}^3)$

Finally, we multiply by $\pi$ to obtain the volume:

$V ≈ 3.14159 * 980 \text{ m}^3$

$V ≈ 3078.76 \text{ m}^3$

Therefore, the cylindrical container can hold approximately 3078.76 cubic meters of water. This calculation demonstrates the practical application of geometric formulas in determining the capacity of real-world objects. The concept of volume is crucial in various fields, including engineering, construction, and manufacturing. Understanding how to calculate the volume of different shapes allows professionals to design structures, estimate material requirements, and optimize storage solutions. The unit of volume, cubic meters ($m^3$), represents the amount of space occupied by a three-dimensional object. It's important to use consistent units throughout the calculation to ensure accurate results. In this case, we used meters for both the radius and the height, resulting in a volume expressed in cubic meters. Furthermore, this calculation highlights the importance of the mathematical constant $\pi$. This irrational number, which represents the ratio of a circle's circumference to its diameter, appears in numerous geometric formulas and is essential for accurate calculations involving circular shapes. The application of the cylinder volume formula extends beyond just water storage. It can be used to calculate the volume of any cylindrical object, such as pipes, tanks, and even food containers. This versatility makes it a valuable tool in various industries. In addition to calculating the volume, understanding the surface area of a cylinder is also important in many applications. The surface area determines the amount of material needed to construct the cylinder or the amount of coating required to cover it. The formula for the surface area of a cylinder is:

$A = 2\pi r^2 + 2\pi rh$

where the first term represents the area of the two circular bases, and the second term represents the lateral surface area. By mastering the calculation of both volume and surface area, one can gain a comprehensive understanding of the geometric properties of cylinders. The ability to calculate the volume of a cylinder is a fundamental skill that has practical applications in various fields. From determining the capacity of storage tanks to designing cylindrical structures, this knowledge is essential for engineers, architects, and other professionals. By understanding the formula and applying it correctly, you can confidently tackle real-world problems involving cylindrical shapes. Geometry is more than just abstract shapes and formulas; it's a powerful tool for understanding and interacting with the physical world. The calculations we've performed here are just one example of how geometric principles can be applied to solve practical problems. By continuing to explore the world of geometry, you'll discover its beauty and its power to shape our understanding of the universe. This example clearly shows that mathematical concepts are not confined to textbooks and classrooms; they are integral to our daily lives and play a significant role in various industries. By connecting mathematical knowledge with real-world applications, we can appreciate its value and its potential to solve problems and improve our world.