Problem 15 - Precalculus

Question

Use the property: \(b^{a}=c\) if and only if \(\log _{b}(c)=a\) from Theorem 6.2 to rewrite the given equation in the other form. That is, rewrite the exponential equations as logarithmic equations and rewrite the logarithmic equations as exponential equations. \(\ln \left(\frac{1}{\sqrt{e}}\right)=-\frac{1}{2}\)

Step-by-Step Solution

Answer

The equation is rewritten in exponential form as \(e^{-\frac{1}{2}} = \frac{1}{\sqrt{e}}\).

Step 1: Identify the Base

The term \(\ln\) indicates a natural logarithm, which has a base of \(e\). So, we recognize that the logarithmic form given \(\ln \left(\frac{1}{\sqrt{e}}\right)=-\frac{1}{2}\) can be expressed with base \(e\) in an exponential form.

Step 2: Set Up the Exponential Form

Using the theorem \(b^{a}=c\) if and only if \(\log _{b}(c)=a\), we have \( \ln(c) = a\), which means \(e^{a} = c\). Here, \(c = \frac{1}{\sqrt{e}}\) and \(a=-\frac{1}{2}\).

Step 3: Write the Equation in Exponential Form

Rewrite the equation \(\ln \left(\frac{1}{\sqrt{e}}\right)=-\frac{1}{2}\) as an exponential equation: \[ e^{-\frac{1}{2}} = \frac{1}{\sqrt{e}} \].

Key Concepts

Natural Logarithms

Natural logarithms are a specific type of logarithm that use the base of the mathematical constant \(e\), approximately equal to 2.71828. This base \(e\) is a unique number that appears frequently in mathematics, particularly in calculus. When you see \(\ln(x)\), it implies \(\log_e(x)\).

Natural logarithms are crucial for solving equations involving exponential growth and decay, such as population growth models or half-life calculations in chemistry. They give you the power to "undo" exponentials with base \(e\), making them a valuable tool for simplifying complex expressions. \(\ln(x)\) tells you what power to raise \(e\) to, in order to get \(x\).

Using natural logarithms is especially helpful when equations cannot be easily solved through simple algebraic manipulations. For instance, if you stumble upon an equation like \(e^x = 20\), converting it into a logarithmic form with \(\ln\) can make it much easier to solve.

Logarithmic Identities

Logarithmic identities are essential for simplifying and solving logarithmic equations. They make it easier to manipulate expressions and calculate results. Here are some key identities to remember:

• \(\log_b(1) = 0\) because any number raised to the power of 0 is 1.
• \(\log_b(b) = 1\) because any base raised to the power of 1 is itself.
• \(\log_b(mn) = \log_b(m) + \log_b(n)\) - This shows how logarithms convert multiplication into addition.
• \(\log_b(\frac{m}{n}) = \log_b(m) - \log_b(n)\) - Similarly, logarithms turn division into subtraction.
• \(\log_b(m^n) = n \cdot \log_b(m)\) - This rule helps in dealing with powers inside a logarithm.

These identities are instrumental when rewriting logarithmic expressions or when switching between logarithmic and exponential forms, such as during processes of solving exponential and logarithmic equations. Understanding these principles of logarithms simplifies how you approach and solve various mathematical problems.

Exponential Identities

Exponential identities are fundamental in understanding how exponential growth, decay, and transformations operate. They help us write and comprehend equations involving exponential quantities. Consider these major exponential rules:

• \(a^0 = 1\) for any non-zero \(a\).
• \(a^{-n} = \frac{1}{a^n}\)
• \((a^m)^n = a^{mn}\)
• \(a^{m+n} = a^m \cdot a^n\)

Exponential identities often come in handy when solving equations or rewriting expressions. For instance, converting logarithmic equations into exponential form uses these identities.

Consider the original exercise we've been working with: \(\ln \left(\frac{1}{\sqrt{e}}\right) = -\frac{1}{2}\). By understanding that \(\ln(x)\) is the logarithm to the base \(e\), we can apply our knowledge of the identity \(e^a = c\) if \(\ln(c) = a\) to deduce \(e^{-\frac{1}{2}} = \frac{1}{\sqrt{e}}\). This is a straightforward application of both logarithmic and exponential identities in finding solutions.