Question

The logarithm of the quotient of two numbers is equal to the difference of the logarithms of the numbers.

Short answer

\(\log(10/2) = 0.7\) and \(\log(10) - \log(2) = 0.699\). These results showcase the property that the log of the quotient of two numbers equals the difference of logs of those numbers independently, with minimal discrepancy due to the approximation of logarithmic values.

Step-by-step solution

Understand The Given Statement

The exercise states that the logarithm of the quotient of two numbers equals the difference of the logarithms of the numbers. In mathematical terms, if we have two numbers \(a\) and \(b\), this can be expressed as: \(\log(a/b)=\log(a)-\log(b)\). This is a general rule of logarithms.

Formulating An Example

To further illustrate this concept, let's take specific numbers. For example, let's consider \(a = 10\) and \(b = 2\). Thus, we have \(\log(10/2) = \log(10) - \log(2)\).

Execute The Logarithmic Calculations

Calculating the left side will give us \(\log(5)\). Calculating the right side using a logarithms calculator (as logarithms are typically not calculated by hand), gives us \(\log(10) - \log(2) = 1 - 0.301 = 0.699\). These two values will not be equivalent due to the approximation of the logarithmic values. Therefore, this discrepancy is expected and is seen in many mathematical exercises involving logarithms.

Key concepts

Logarithm of a Quotient

Understanding the logarithm of a quotient is crucial for solving various mathematical problems, particularly those involving logarithmic expressions. The exercise statement presents us with an important logarithmic property that can be stated as: When you have two numbers, say a and b, and you want to find the logarithm of their quotient, you can simply find the difference of their individual logarithms. Mathematically, this is expressed as \( \log\left(\frac{a}{b}\right) = \log(a) - \log(b) \).

For instance, if you have \( \frac{100}{10} \), instead of directly calculating \( \log(10) \), you could calculate \( \log(100) - \log(10) \), which simplifies to \( 2 - 1 = 1 \), saving time and effort in the process.

Logarithmic Calculations

When it comes to logarithmic calculations, precision and understanding of logarithmic tables or calculators are key. Calculating logarithms directly by hand is a skill seldom used today, as we typically rely on calculator tools for accurate results. When we calculated \( \log(10) - \log(2) \) in the example, we used a calculator because getting the exact logarithmic values of these numbers is not straightforward.

It's essential to remember that the results of logarithmic calculations on a calculator are often rounded approximations. This means they may not be exact, but for most practical purposes, they are sufficiently precise. Therefore, if you obtain \( \log(5) \) as \( 0.699 \), this is an expected approximation for the actual logarithmic value.

Rules of Logarithms

There are several rules of logarithms that one must grasp to manipulate and solve logarithmic expressions effectively. Besides the logarithm of a quotient, there are other significant properties such as:

• Product Rule: \( \log(ab) = \log(a) + \log(b) \), meaning the logarithm of a product is the sum of the logarithms.
• Power Rule: \( \log(a^n) = n \cdot \log(a) \), which tells us that the logarithm of a power is the exponent times the logarithm of the base.
• Change of Base Formula: \( \log_b(a) = \frac{\log_c(a)}{\log_c(b)} \), useful for converting logarithms from one base to another.
• Logarithm of One: \( \log(1) = 0 \), since any number to the power of 0 is 1.
• Logarithm of the Base: \( \log_b(b) = 1 \), because any base raised to the power of 1 is the base itself.

Understanding and applying these rules can greatly simplify logarithmic problem-solving and are essential for algebraic manipulation involving logarithms.

Logarithmic Equations

Solving logarithmic equations requires understanding logarithmic properties and applying them methodically. Logarithmic equations contain logarithms with variables that you are required to solve for. The key to solving these equations is often to use the logarithmic rules to 'condense' or 'expand' the logarithmic expressions to a form where the variables can be isolated and made the subject of the formula.

For example, if an equation is given as \( \log(x+1) - \log(2) = 1 \), you can first use the property involving the logarithm of a quotient to combine the terms into a single logarithm. Assuming a common base of 10, this would become \( \log\left(\frac{x+1}{2}\right) = 1 \), which can then be transformed into an exponential equation: \( 10^1 = \frac{x+1}{2} \). Solving for \( x \) will give you the unknown value. This method shows how understanding logarithmic properties directly leads to the ability to solve equations involving logarithms.