Problem 92 - Introductory and Intermediate Algebra for College Students 4th

Question

The loudness level of a sound, \(D,\) in decibels, is given by the formula $$D=10 \log \left(10^{12} I\right)$$ where I is the intensity of the sound, in watts per meter \(^{2} .\) Decibel levels range from \(0,\) a barely audible sound, to \(160,\) a sound resulting in a nuptured eardrum. Use the formula to solve Exercises. What is the decibel level of a normal conversation, \(3.2 \times 10^{-6}\) watt per meter \(^{2} ?\)

Step-by-Step Solution

Answer

The decibel level of a normal conversation that has an intensity of \(3.2 \times 10^{-6}\) watts per meter squared is approximately 65.05 decibels.

Step 1: Understand the formula

The given formula is \(D=10 \log \left(10^{12} I\right)\), where \(D\) is the loudness level to find and \(I\) is the known intensity of the sound in watts per meter squared. In this case, \(I = 3.2 \times 10^{-6}\) watt per meter squared.

Step 2: Substitute the value of I into the formula

Substitute the given intensity \(I = 3.2 \times 10^{-6}\) watts per meter squared into the formula. This gives: \(D=10 \log \left(10^{12} \times 3.2 \times 10^{-6}\right)\)

Step 3: Simplify the expression

Next, simplify the numbers inside the logarithm using the property of logarithms: \(\log \left(10^{12} \times 3.2 \times 10^{-6}\right) = \log \left(3.2 \times 10^{6}\right)\). Now, substitute this into the formula: \(D=10 \log \left(3.2 \times 10^{6}\right)\)

Step 4: Evaluate the Logarithm

Next, calculate the logarithm. Note that the base of logarithm is 10, therefore \(\log \left(3.2 \times 10^{6}\right)\) can be separated using the property of logarithms into \(\log 3.2 + \log \left(10^{6}\right)\). Since \(\log \left(10^{6}\right) = 6\), the formula simplifies to \(D=10 \(6+\log 3.2\)\)

Step 5: Final Calculation

Finally, perform the arithmetic to obtain the decibel level \(D = 10 \(6+\log 3.2\)\). Use a calculator to find that \(\log 3.2 \approx 0.5051\), thus the final decibel level is \(D = 10 \(6 + 0.5051\) \approx 65.05\) decibels.

Key Concepts

Logarithms in Algebra

Logarithms play a significant role in solving various algebraic problems and are pivotal in understanding the calculation of decibel levels for sounds. The logarithm of a number is the exponent to which a fixed number, the base, must be raised to produce that number. In our context, the logarithm has a base of 10, which is common in scientific calculations.

Properties of Logarithms

When calculating decibel levels, we utilize the property that the logarithm of a product is the sum of the logarithms (i.e., \(\log(ab) = \log(a) + \log(b)\)). This allows us to simplify multiplication inside the logarithm into more manageable terms. For instance, in decibel calculations, this property enables us to separate the intensity level of the sound and the reference intensity. Understanding how to manipulate logarithms in algebra is essential for accurate and efficient sound level calculations.

Intensity of Sound

The intensity of sound is a measure of the power per unit area, expressed in watts per square meter (W/m\(^2\)). It's a crucial concept to understand as it quantifies the amount of sound energy passing through a given area. In acoustics, sound intensity is not only important for measuring loudness but also for sound insulation and engineering purposes.

Understanding Sound Intensity

Sound intensity levels vary depending on the environment and the source. For instance, a whisper might be around 0.000001 W/m\(^2\), whereas a jet engine can be upwards of 100 W/m\(^2\). When plugged into the decibel formula, larger intensities yield higher decibel levels, hence louder sounds. For students and professionals alike, grasping this concept is fundamental when studying sound and its properties.

Decibels in Acoustics

Decibels, abbreviated as dB, are a logarithmic unit used to measure the intensity of sound. In acoustics, this measurement is instrumental in comparing different sound levels. The decibel scales logarithmically; thus, it can encompass the very wide range of sound intensities humans can hear, from the faintest detectable to those that can cause physical pain or even damage.

How Decibels Work

The relationship between intensity and perceived loudness isn't linear—which means doubling the intensity doesn't necessarily double the perceived loudness. Hence, a logarithmic scale like decibels becomes invaluable as it reflects the way our ears perceive changes in loudness. Learning about decibels in acoustics not only helps students in technical fields like sound engineering but also is relevant to everyday experiences, such as understanding the potential hearing damage from loud music.

Solving Algebraic Expressions

In the context of decibel level calculations, solving algebraic expressions becomes essential. It involves manipulating and simplifying those expressions according to algebraic rules and properties, such as the distributive property, associative property, and logarithmic properties.

Simplification Techniques

To simplify the process, we can break down complex expressions into simpler parts using these rules. For instance, in our exercise, the exponent property of logarithms (\(\log(a^b) = b\log(a)\)) is used to evaluate the logarithmic part of the formula. This property greatly simplifies the process, transforming a potentially challenging multiplication into a manageable addition problem. Crucial for students to master, these techniques streamline the path to finding the solution to algebraic problems, especially in physical applications like acoustics.