Problem 5 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

For Problems 3 through 9 , simplify the expression given. (a) \(10^{\log 2+1}\) (b) \(e^{3-\ln 2}\)

Step-by-Step Solution

Answer

The simplified forms of the expressions \(10^{\log 2+1}\) and \(e^{3-\ln 2}\) are 20 and \(\frac{e^3}{2}\) respectively.

Step 1: Simplify the expression \(10^{\log 2+1}\)

Here, we will use the base property of logarithms which states that if b^z = y then z = log_b y. In this case, we can rewrite the expression by taking '2' as base 10 and '1' as the power of 10: \(10^{\log_{10} 2 + log_{10} 10} = 10^{\log_{10} (2 * 10)} = 10^{\log_{10} 20}\). Now, using the property we mentioned , 10 to the log base 10 of 20 = 20.

Step 2: Simplify the expression \(e^{3-\ln 2}\)

Here, we will use the similar base property of natural logarithms (\(ln\)) , that if e^z = y then z = ln y , and also the laws of exponents for real numbers. We can rewrite the expression by taking '3' as power of e and '-2' as the natural log base e : \(e^{ln e^3 - ln 2} = e^{ln \frac{e^3}{2}} \). Because, \(e^{ln y} = y\), we can simplify the expression to: \(e^{ln \frac{e^3}{2}} = \frac{e^3}{2} \).

Key Concepts

Simplifying Expressions

Simplifying algebraic expressions is an essential skill in mathematics, particularly when dealing with exponents and logarithms. The main goal is to make the expressions easier to understand and work with, by applying various properties and rules to break down complex terms into simpler forms.

Take for example the expression 10 to the power of \(\log 2+1\). Simplifying such an expression involves recognizing and applying logarithmic properties. Since \(10^{\log_{10} x} = x\), we can see \(\log 2 + 1\) as \(\log_{10} 2 + \log_{10} 10\). This transforms our expression into 10 to the power of the \(\log_{10}\) of 20, which then simplifies directly to 20, because 10 raised to the logarithm of a number to the base 10 is the number itself.

Always remember, the key to simplification is identifying patterns and properties that can be applied; whether breaking down an exponent using the power rule or combining logarithmic terms with the same base using the product rule for logarithms.

Exponents and Logarithms

Exponents and logarithms are two sides of the same mathematical coin, with logarithms effectively acting as the inverse operation of exponentiation. When simplifying expressions involving these concepts, it's crucial to understand this relationship.

For instance, the expression 10 raised to the \(\log 10\) power represents the inverse process of raising 10 to a power and then taking the logarithm of the result. This duality means that \(10^{\log_{10} x}\) is simply \(x\), an application of the inverse property that simplifies the expression greatly.

Similarly, for natural logarithms and \(e\) (Euler's number), the relationship is such that \(e^{\ln x} = x\) and \(\ln(e^x) = x\). This connection allows for simplifications of expressions like \(e^{3 - \ln 2}\), which can be rephrased using logarithmic subtraction as \(e^{\ln e^3 - \ln 2}\), or \(e^{\ln (e^3/2)}\), which simplifies to \(e^3/2\).

Natural Logarithms

The natural logarithm, denoted as \(\ln\), uses the number \(e\) (approximately 2.71828) as its base. It is widely used in various fields of science and mathematics due to its natural properties in describing growth processes and compound interest.

Let's delve into an example. Taking the expression \(e^{3 - \ln 2}\), our first step is to dissect it using properties of natural logarithms. We recognize that \(\ln a - \ln b = \ln (a/b)\) which implies that \(e^{3 - \ln 2}\) can be rewritten as \(e^{\ln (e^3/2)}\). In the spirit of simplification, we then apply the fundamental property that \(e^{\ln x} = x\), thus the expression boils down to \(e^3/2\).

Understanding natural logarithms and their properties allows for these simplifications, transforming what might seem like complex expressions into intuitive and manageable solutions.