Problem 83 - Calculus for Biology and Medicine

Question

Write the following expressions in terms of base \(e\), and simplify: (a) \(3^{x}\) (b) \(4^{x^{2}-1}\) (c) \(2^{-x-1}\) (d) \(3^{-4 x+1}\)

Step-by-Step Solution

Answer

(a) \(e^{x \ln(3)}\); (b) \(e^{x^2 \ln(4) - \ln(4)}\); (c) \(e^{-x \ln(2) - \ln(2)}\); (d) \(e^{-4x \ln(3) + \ln(3)}\)

Step 1: Understand the Conversion to Base e

To convert an expression with base other than \(e\) to base \(e\), use the property of logarithms \(a^{x} = e^{x \ln(a)}\). We'll apply this property to each expression.

Step 2: Convert \(3^x\) to Base e

Using the property \(a^x = e^{x \ln(a)}\), convert \(3^x\) as follows: \[3^x = e^{x \ln(3)}\]

Step 3: Convert \(4^{x^2 - 1}\) to Base e

Apply the conversion to base \(e\): \[4^{x^2 - 1} = e^{(x^2 - 1) \ln(4)}\]

Step 4: Convert \(2^{-x-1}\) to Base e

Again, use the logarithm property: \[2^{-x-1} = e^{(-x-1) \ln(2)}\]

Step 5: Convert \(3^{-4x+1}\) to Base e

Finally, convert the expression: \[3^{-4x+1} = e^{(-4x+1) \ln(3)}\]

Step 6: Simplify using Distribution and Common Laws

For each expression, if possible, distribute the multiplication inside the exponent to simplify: - For \(3^x\), it is already simplified as \(e^{x \ln(3)}\).- For \(4^{x^2 - 1}\), simplify to \(e^{x^2 \ln(4) - \ln(4)}\).- For \(2^{-x-1}\), simplify to \(e^{-x \ln(2) - \ln(2)}\).- For \(3^{-4x + 1}\), simplify to \(e^{-4x \ln(3) + \ln(3)}\).

Key Concepts

Natural Logarithm

The natural logarithm is a powerful tool in mathematics. It is denoted as \( \ln \) and represents the logarithm to the base \(e\), which is approximately equal to 2.71828. The natural logarithm is particularly useful when dealing with exponential functions, especially when we want to rearrange expressions for various calculations or simplifications.

Using the property \( a^x = e^{x \ln(a)} \), we can convert any exponential expression to have the base \( e \). This transformation makes complex algebraic manipulation more manageable. Natural logarithms allow us to work within the framework of the well-known exponential growth and decay models, which simplifies problems across mathematics and science.

When you see an expression like \( 3^x \), you can transform it to base \( e \) by noting that \( 3^x = e^{x \ln(3)} \). Here, \( \ln(3) \) is a constant that provides a multiplier for the exponent, effectively translating the original base 3 exponential into a form you're likely to work with in calculus and differential equations.

Base Conversion

Base conversion is a technique used to express any exponential expression in terms of another base. It is particularly handy in calculus and algebra for simplifying complex expressions. By using the rule \( a^x = e^{x \ln(a)} \), you can convert any base exponential expression to base \( e \). This rule taps into the natural properties of \( e \), making it easier to solve equations or differentiate and integrate functions.

• For expression \( 4^{x^2 - 1} \), converting it involves using \( \ln(4) \), leading to \( e^{(x^2 - 1) \ln(4)} \).
• The expression \( 2^{-x-1} \) converts to base \( e \) as \( e^{(-x-1) \ln(2)} \).
• Similarly, \( 3^{-4x+1} \) becomes \( e^{(-4x+1) \ln(3)} \).

Base conversion sets expressions into a standard form, simplifying operations like differentiation and integration. It's crucial to remember that \( \ln(a) \) is just a constant factor that adjusts the exponent when transformation occurs. This constant is what makes base \( e \) such a universal choice when handling exponential functions.

Simplification of Expressions

Simplifying expressions is a central task in algebra and calculus. It involves rewriting an expression in the simplest form possible while maintaining its original value. After converting an expression to base \( e \) using natural logarithms, the next step involves simplifying it using distribution and laws of exponents.
To simplify expressions like the ones we converted to base \( e \):

• For \( 3^x \), we noted it was already simplified as \( e^{x \ln(3)} \).
• The expression \( 4^{x^2-1} \) simplifies to \( e^{x^2 \ln(4) - \ln(4)} \) by distributing \( \ln(4) \) over the terms inside the exponent.
• For \( 2^{-x-1} \), the simplification leads to \( e^{-x \ln(2) - \ln(2)} \), clearly dividing the exponent into two separate terms.
• In the case of \( 3^{-4x+1} \), it simplifies to \( e^{-4x \ln(3) + \ln(3)} \), distributing the logarithmic term accordingly.

Simplifying exponential expressions helps streamline calculations and predictions about system behavior, especially when executed under rules of calculus. Each part of the expression can reveal how different terms affect the function’s growth or decay.