Problem 83 - Precalculus : Building Concepts and Connections

Question

The decibel (dB) is a unit that is used to express the relative loudness of two sounds. One application of decibels is the relative value of the output power of an amplifier with respect to the input power. since power levels can vary greatly in magnitude, the relative value \(D\) of power level \(P_{1}\) with respect to power level \(P_{2}\) is given (in units of \(\mathrm{dB}\) ) in terms of the logarithm of their ratio as follows: $$D=10 \log \frac{P_{1}}{P_{2}}$$ where the values of \(P_{1}\) and \(P_{2}\) are expressed in the same units, such as watts \((\mathrm{W}) .\) If \(P_{2}=75 \mathrm{W},\) find the value of \(P_{1}\) at which \(D=0.7\)

Step-by-Step Solution

Answer

Upon solving the above equation, you find that the value of \(P_{1}\) is approximately 79.43W.

Step 1: Substitute provided values

Firstly, substitute the provided values of D and \(P_{2}\) into given formula: \(0.7 = 10 \log \frac{P_{1}}{75}\)

Step 2: Isolate the logarithmic term

Secondly, to isolate the logarithmic term on one side, divide both sides by 10 : \(0.07 = \log \frac{P_{1}}{75}\)

Step 3: Transform the logarithmic equation to exponential form

The base of the logarithm is 10 (since it is not specified). Thus, the equation from step 2 can be written in exponential form as: \(10^{0.07} = \frac{P_{1}}{75}\)

Step 4: Solve for \(P_{1}\)

Multiply both sides by 75 to solve for \(P_{1}\): \(P_{1} = 75 * 10^{0.07}\)

Key Concepts

Understanding Decibels

Decibels (dB) are used as a unit for measuring sound intensity levels. While "intensity" is a scientific term, simply think of it as "loudness". However, decibels are not only for sound. They are widely used in electronics and communications to express the ratio of two power levels. The primary reason for using decibels is because they can represent very large or very small numbers in a convenient manner thanks to logarithms.

Consider why decibels are so effective:

• They provide a relative measure, making it easier to compare different power levels.
• They compress the scale of measurements, which is quite useful since power levels in electronics can vary drastically.
• They are based on a logarithmic scale, which simplifies certain types of mathematical calculations, such as multiplying power ratios.

In the context of this exercise, understanding decibels helps in analyzing the change in power when the input and output levels of an amplifier are compared.

Exploring the Power Ratio

The power ratio is a comparison between two quantities of power. In the world of physics and engineering, it's essential whenever you want to know how much power gain or loss occurs, such as through devices like amplifiers.

Consider the formula for power levels in decibels:\[ D = 10 \, \log \frac{P_1}{P_2} \]Here,

• \( P_1 \) is the output power.
• \( P_2 \) is the input power.
• \( D \) is the decibel value representing the power ratio.

This mathematical expression shows that if \( P_1 \) equals \( P_2 \), the ratio is \( 1 \), and the logarithm of \( 1 \) is \( 0 \). So, \( D \) would be zero, indicating no power loss or gain. Meanwhile, when \( D \) is positive, it indicates a power gain, and if negative, a power loss.

Working with Exponential Equations

Exponential equations are a staple in mathematics, involving terms where variables are present in the exponent. They frequently appear in problems dealing with growth rates, decay, and in this case, calculations with logarithms and decibels.

When solving this exercise, we need to convert a logarithmic equation to an exponential form:

• You start with a logarithmic equation: \( \log \frac{P_1}{75} = 0.07 \).
• Remember that logarithms are the inverse of exponents, which means you can "undo" a log by raising the base (in this case, 10) to the power of both sides of the equation.
• This conversion results in an exponential equation: \( 10^{0.07} = \frac{P_1}{75} \).
• Solving this yields an answer for \( P_1 \), the unknown power, which we achieve through simple algebraic manipulation.

Understanding how to shift between these forms is invaluable. It enhances your ability to solve a wide array of mathematical problems efficiently.