Problem 4 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

For Problems 3 through 9 , simplify the expression given. (a) \(3^{2} 10^{2 \log 5}\) (b) \(5 e^{-3 \ln 2}\)

Step-by-Step Solution

Answer

(a) The simplified form of \(3^{2} 10^{2 \log 5}\) is 225. (b) The simplified form of \(5 e^{-3 \ln 2}\) is 5/8.

Step 1: Simplify the first expression

Using the property a^log(a^b) = a^b, we can simplify the expression \(3^{2} 10^{2 \log 5}\) as follows:(a) \(3^{2} * 10^{2 \log 5} = 9 * 10^{\log 25} = 9 * 25 = 225.\)

Step 2: Simplify the second expression

First, we can use the logarithmic power rule, log(a^m) = m log(a), to simplify the exponential form e^{-3 \ln 2}. Then, use the property e^(-x) = 1/e^x to simplify the expression:(b) \(5 * e^{-3 \ln 2} = 5 * e^{\ln (2^{-3})} = 5 * 2^{-3} = 5/8.\)

Key Concepts

Exponential Functions: Understanding Growth and Decay

Exponential functions are mathematical expressions where a constant base is raised to a variable exponent. They are crucial in modelling numerous real-world phenomena such as population growth, radioactive decay, and interest calculations. In general, an exponential function can be written in the form \(a^x\), where \(a\) is a positive constant called the base and \(x\) represents the exponent. Here are a few properties of exponential functions:

• Rapid Growth or Decay: Depending on the base value, exponential functions can model rapid growth (when \(a > 1\)) or decay (when \(a < 1\)).
• Base Conversion: If different bases are involved, converting them using logarithms can simplify the analysis. This is often done through the change of base formula.

For solving exponential expression, like the original exercise part (b) \(5 e^{-3 \ln 2}\), recognizing that \(e^{\ln (2^{-3})}\) simplifies to \(2^{-3}\) by using the basic property \(a^{\ln b} = b\) for natural logarithms, and results in a simplified expression \(5 \times (\frac{1}{8}) = \frac{5}{8}\). This showcases how exponential functions can be manipulated for simplification.

Logarithmic Properties: The Inverse of Exponentials

Logarithms are the inverses of exponential functions, meaning they help us "undo" exponentiation. If you have an expression involving exponentials and want to solve it or simplify it, logarithms are your go-to tool. The general rule of logarithms is that \(\log_b(a) = c\) means that \(b^c = a\). Logarithms have several powerful properties that make them especially useful:

• Product Property: \(\log_b(xy) = \log_b(x) + \log_b(y)\)
• Quotient Property: \(\log_b(\frac{x}{y}) = \log_b(x) - \log_b(y)\)
• Power Property: \(\log_b(x^m) = m \cdot \log_b(x)\), which helps in simplifying powers.

In the original exercise, these properties, especially the power property, were critical for simplification. Part (a) of the exercise used the property \(a^{\log_b(a^k)} = k\), which simplifies the expression \(10^{2 \log 5}\) to \(25\), because it utilizes the identity \(b^{\log_b x} = x\) in simplifying to the expression \(10^{\log 25}\). These transformations show how handy logarithmic properties are for restructuring complex expressions.

Mathematical Simplification: Making Expressions Manageable

Mathematical simplification involves rewriting expressions in a more compact or more easily understandable form. Simplification can help you see relationships between components of an equation or reduce errors in subsequent computations. There are different strategies and tools we use to simplify expressions:

• Using Properties: Utilize mathematical properties like commutative, associative, and distributive laws for arithmetic operations.
• Combining Like Terms: Grouping similar terms to simplify algebraic expressions.
• Canceling Common Factors: When working with fractions or division, reduce by cancelling common factors from the numerator and denominator.
• Applying Function Properties: By using the properties of fundamental functions like exponentials and logarithms, expressions can often be dramatically reduced.

For instance, in the expression \(3^{2} \cdot 10^{2 \log 5}\), recognizing that \(10^{\log 25}\) simplifies directly to \(25\) using properties of logarithms allows us to calculate it directly as \(9 \cdot 25 = 225\). Simplification makes complex expressions more approachable and computation-friendly, showing you the results in a clear and concise manner.