Problem 3 - Algebra and Trigonometry

Question

Evaluate the expression. $$ \log 4+\log 25 $$

Step-by-Step Solution

Answer

The value of the expression is 2.

Step 1: Use the Property of Logarithms

We can use the property of logarithms, which states that \( \log a + \log b = \log (ab) \). This property allows us to combine the two logarithmic terms into a single logarithm with a product inside.

Step 2: Apply the Property

Apply the property from Step 1 to the given expression: \[ \log 4 + \log 25 = \log (4 \times 25) \] This simplifies the expression by combining the terms inside the logarithm.

Step 3: Simplify the Product

Calculate the product inside the logarithm: \( 4 \times 25 = 100 \). This simplifies to: \( \log (4 \times 25) = \log 100 \).

Step 4: Evaluate the Logarithm

Now, evaluate \( \log 100 \). Since \( 100 = 10^2 \), and knowing that the base of the logarithm is 10, we use the fact that \( \log_{10} (10^n) = n \): \( \log 100 = \log (10^2) = 2 \).

Key Concepts

Properties of Logarithms

Logarithms have special properties that make them very powerful in simplifying expressions and solving equations. One key property is the product property. This states that the logarithm of a product is equal to the sum of the logarithms of the individual factors. Mathematically, this is expressed as

• \( \log a + \log b = \log(ab) \)

This property allows us to transform addition of logarithms into a single logarithm with a multiplied argument inside.
For example, if you have \( \log 2 + \log 5 \), you can combine them into \( \log(2 \times 5) = \log 10 \). It’s particularly useful when you're faced with complex logarithmic expressions.
Understanding these properties helps in simplifying calculations and solving problems more effectively!

Evaluating Logarithmic Expressions

When evaluating logarithmic expressions, understanding the concept of base and power is crucial. To evaluate expressions like the one in the exercise, \( \log 4 + \log 25 \), the product property of logarithms becomes handy. This simplifies the evaluation process considerably by reducing the number of terms.
In the given expression:

• Apply the product property to get \( \log(4 \times 25) \).
• Calculate the product: \( 4 \times 25 = 100 \).
• So, you're left with \( \log 100 \).

Knowing the base of the logarithm here is 10 (the common or decimal logarithm), recognize that \( 100 = 10^2 \). From here, since \( \log_{10}(10^n) = n \), you can conclude that:

• \( \log 100 = \log(10^2) = 2 \).

This results in a straightforward answer and demonstrates the simplicity brought by using logarithmic properties.

Logarithmic Functions

Logarithmic functions are the inverse functions of exponential functions. Essentially, if you know how exponents work, understanding logarithms becomes much easier. These functions help in finding the power to which a given base must be raised to obtain a certain number.
In simpler terms, if you have \( y = \log_b(x) \), it means that \( b^y = x \). Here, \( b \) is the base, \( y \) is the logarithm, and \( x \) is the argument or the number you're evaluating. For example, \( \log_{10}(100) = 2 \) represents the fact that \( 10 \) raised to the power of \( 2 \) equals \( 100 \).
These functions are prevalent in various fields, including science and engineering, for modeling growth processes and solving equations involving exponentiation. The common (base-10) and natural (base \( e \)) logarithms are the most frequently used, each having its own significance depending on the application.

• Common logarithms (base-10) are used for measures of sound intensity in decibels or in pH calculations for acidity.
• Natural logarithms (base \( e \)) frequently appear in continuous growth models like population growth and radioactive decay.

Understanding the characteristics of logarithmic functions is essential to effectively apply their properties and solve real-world problems.