Problem 2 - Basic Engineering Mathematics

Question

Evaluate (a) lg 0.001 (b) lne (c) \(\log _{3} \frac{1}{81}\)

Step-by-Step Solution

Answer

(a) -3, (b) 1, (c) -4.

Step 1: Evaluate lg 0.001

To find the common logarithm (base 10) of 0.001, we use the property \[\log_{10}a^b = b \cdot \log_{10}a\].Here, 0.001 can be written as \(10^{-3}\). Therefore, \[\log_{10}(0.001) = \log_{10}(10^{-3}) = -3 \cdot \log_{10}(10) = -3 \cdot 1 = -3\].

Step 2: Evaluate ln e

The natural logarithm (base e) of e is 1 by definition, i.e., \[\ln(e) = 1\]. This is because e is the base of natural logarithms, meaning \[e^1 = e\].

Step 3: Evaluate \(\log_{3} \frac{1}{81}\)

First, express \(\frac{1}{81}\) as a power of 3. We know 81 is \(3^4\), so \(\frac{1}{81} = 3^{-4}\). Then, apply the property of logarithms: \[\log_{3}(3^{-4}) = -4 \cdot \log_{3}(3) = -4 \cdot 1 = -4\],since \(\log_{3}(3) = 1\).

Key Concepts

Common Logarithms

Common logarithms are logarithms that have a base of 10. They are often written as "log" or "lg" without specifying the base, as base 10 is understood. Common logarithms are particularly helpful because they easily relate to decimal numbers, allowing for a simpler calculation when dealing with powers of ten.
Consider the problem of evaluating the common logarithm of 0.001. It's essential to know that 0.001 can be expressed as a power of ten, specifically, it is equal to \(10^{-3}\).
When you take the logarithm of a power of 10, such as 0.001, the property \(\log_{10} a^b = b \cdot \log_{10} a\) can be used effectively.
Once transformed into \(10^{-3}\), the result is straightforward to compute:

• \(\log_{10}(10^{-3}) = -3 \cdot \log_{10}(10) = -3 \cdot 1 = -3\)

This method shows how common logarithms provide a useful tool for simplifying calculations involving powers of ten.

Natural Logarithms

Natural logarithms differ from common logarithms because they use the constant \(e\) as their base. The symbol for natural logarithms is "ln".
The number \(e \approx 2.718\) is a fundamental constant in mathematics, much like \(\pi\), and it arises in situations involving growth and decay.
A key property of natural logarithms is that the logarithm of \(e\) itself is always 1. This happens because \(e\) is the base, and thus \(\ln(e)\) asks "how many times must we multiply \(e\) to get \(e\)?" The answer is precisely 1, as \(e^1 = e\). This makes computing natural logarithms of \(e\) very simple:

• \(\ln(e) = 1\)

Understanding the natural logarithm is crucial as it appears frequently in calculus, especially when dealing with continuous growth, such as in problems involving compound interest or population growth.

Properties of Logarithms

The properties of logarithms make it easier to manipulate and calculate expressions involving logs. Several properties emerge regularly and simplify complex logarithmic expressions:

• Product Property: \(\log_b(mn) = \log_b(m) + \log_b(n)\)


• Quotient Property: \(\log_b\left(\frac{m}{n}\right) = \log_b(m) - \log_b(n)\)


• Power Property: \(\log_b(m^n) = n \cdot \log_b(m)\)


• Change of Base Formula: \(\log_b(a) = \frac{\log_c(a)}{\log_c(b)}\)\, useful when calculators only allow base 10 or base \(e\).

These properties are derived from the definition of logarithms as the inverse of exponentiation. They can be extremely helpful, as seen in evaluating expressions like \(\log_{3}(\frac{1}{81})\). First, recognize \(\frac{1}{81}\) as \(3^{-4}\), then use the Power Property to find:

• \(\log_{3}(3^{-4}) = -4 \cdot \log_{3}(3) = -4 \cdot 1 = -4\)

By applying these properties, seemingly complicated logarithmic expressions become manageable.