Problem 52 - Algebra 2 and Trigonometry

Question

In \(45-52,\) if \(\ln a=c,\) express each of the following in terms of \(c\) $$ \ln \sqrt{a} $$

Step-by-Step Solution

Answer

\(\ln \sqrt{a} = \frac{1}{2} c\)

Step 1: Understand the Problem

We are given that \(\ln a = c\) and we need to express \(\ln \sqrt{a}\) in terms of \(c\). From the problem statement, \(\sqrt{a}\) can be rewritten as \(a^{1/2}\).

Step 2: Apply Logarithmic Rule

We use the logarithm power rule: \(\ln (a^{b}) = b \ln a\). In this case, we have \(b = \frac{1}{2}\). Thus, \(\ln \sqrt{a} = \ln (a^{1/2}) = \frac{1}{2} \ln a\).

Step 3: Substitute Given Value

From the problem, \(\ln a = c\). Substitute this into the expression obtained: \(\ln \sqrt{a} = \frac{1}{2} \ln a = \frac{1}{2} c\).

Key Concepts

Natural Logarithm

The natural logarithm, denoted as \(\ln\), is a logarithm to the base \(e\), where \(e\) is an irrational and transcendental number approximately equal to 2.71828. It is widely used in mathematics due to its natural properties relating to growth and finding derivatives.

• When you see \(\ln\), think of it as asking the question, "To what power must \(e\) be raised to get this number?"
• For example, \(\ln(1) = 0\) because \(e^0 = 1\).
• Similarly, \(\ln(e) = 1\) because \(e^1 = e\).

Natural logarithms are essential in calculus and often used in solving equations involving exponential growth or decay, simplifying the process of differentiation and integration.

Logarithmic Identities

Logarithmic identities are fundamental tools that help manipulate and simplify logarithmic expressions. These identities make it easier to work with complex mathematical problems involving logarithms.

• **Product Rule:** \(\ln(xy) = \ln(x) + \ln(y)\) - allows you to separate the logarithm of a product into a sum of two logs.
• **Quotient Rule:** \(\ln\left(\frac{x}{y}\right) = \ln(x) - \ln(y)\) - breaks down the logarithm of a division into a difference.
• **Power Rule:** \(\ln(x^b) = b\cdot\ln(x)\) - lets you take an exponent in a log to its front as a multiplier.

In our exercise, we used the Power Rule to simplify \(\ln(\sqrt{a})\) to \(\frac{1}{2}\cdot\ln(a)\). These identities are extremely useful for any level of mathematics, especially if you aim to rewrite logarithmic equations in a more workable form.

Properties of Exponents

Understanding exponents is crucial when dealing with logarithms, as they are inversely related. Specifically, a logarithm can be seen as the inverse operation to exponentiation. Here are some essential properties of exponents that frequently come into play:

• **Negative Exponent:** \(x^{-b} = \frac{1}{x^b}\) - an exponent that is negative indicates a reciprocal.
• **Zero Exponent:** \(x^0 = 1\) - any non-zero base raised to the power of zero equals one.
• **Fractional Exponent:** \(x^{1/n} = \sqrt[n]{x}\) - exponentials expressed as fractions represent roots.

These properties are invaluable for simplifying expressions and solving equations involving roots and powers, as seen in the problem where \(\sqrt{a}\) was rewritten as \(a^{1/2}\). This conversion allowed the logarithmic power rule to be applied effectively.