Problem 56 - College Algebra

Question

In Exercises \(53-56,\) rewrite the equation in terms of base \(e\). Express the answer in terms of a natural logarithm and then round to three decimal places. $$y=4.5(0.6)^{x}$$

Step-by-Step Solution

Answer

The equation \( y=4.5(0.6)^{x} \) can be written in terms of a natural logarithm as \( x = -1.956(\ln{(\frac{y}{4.5})}) \)

Step 1: Express in logarithmic form

The equation \( y=4.5(0.6)^{x} \) can be expressed in logarithmic form. Using the definition of logarithm, this can be rewritten as \( x = \log_{0.6}(\frac{y}{4.5}) \)

Step 2: Change to natural logarithmic form

We can apply the change of base formula. This states that for any positive numbers \( a, b \) (not equal to 1), \( \log_{b}{a} \) can be written as \( \frac{\log_{c}{a}}{\log_{c}{b}} \) for any positive value of \( c \). In the case of changing the base to \( e \), \( c = e \). The formula becomes \( x = \frac{\ln{(\frac{y}{4.5})}}{\ln{(0.6)}} \).

Step 3: Evaluate numerical values

The decimal approximation of \( \ln{(0.6)} \) is -0.511. So \( x = \frac{\ln{(\frac{y}{4.5})}}{-0.511} \).

Step 4: Rearrange the equation

Multiplying each side of this equation by -0.511 returns \( x = -1.956(\ln{(\frac{y}{4.5})}) \). This is the required equation.

Key Concepts

Change of Base Formula

When encountering logarithms with different bases, the Change of Base Formula becomes a handy tool. This formula allows us to transform a logarithm from one base to another, specifically to a base that might be more convenient, such as the natural logarithm base, denoted as \( e \). This is particularly useful because natural logarithms are commonly available on scientific calculators and are widely used in mathematical calculations.

The Change of Base Formula is expressed as:

• \( \log_{b}{a} = \frac{\log_{c}{a}}{\log_{c}{b}} \)

Here, \( b \) is the original base, \( a \) is the number you're taking the logarithm of, and \( c \) is the new base you're converting to. In most cases, \( c \) is chosen to be \( e \) to use natural logarithms (\( \ln \)).
In the context of the original exercise, we applied the Change of Base Formula to convert a logarithm of base 0.6 to a natural logarithm. This helped in expressing the equation in terms of \( \ln \), following:

• \( x = \frac{\ln{(\frac{y}{4.5})}}{\ln{(0.6)}} \)

Using this formula simplifies calculations because natural logarithms are easier to evaluate, either manually or using a calculator.

Logarithmic Form

A logarithm is essentially another way of expressing the exponent in an exponential equation.

• For example, in the equation \( y = 4.5(0.6)^{x} \), \( x \) is the exponent that makes \( 4.5(0.6)^{x} \) equal to \( y \).

To express this relationship using a logarithm, we can rewrite it in "logarithmic form". This form is like solving the equation \( b^{x} = a \) by finding \( x = \log_{b}(a) \).

In this exercise, we've rewritten the given equation in logarithmic form:

• \( x = \log_{0.6}(\frac{y}{4.5}) \)

This transformation is helpful because it allows for easier manipulation and understanding of the equation, especially when dealing with exponential relationships.

By structuring the equation this way, we set the stage for applying the Change of Base Formula which is crucial in converting our equation into the desired natural logarithm form.

Decimal Approximation

When working with natural logarithms or any logarithms, certain values are not always straightforward for exact evaluation. This is where decimal approximation comes into play. It's often necessary to convert logarithmic values into decimal form for simpler calculation and interpretation.

In the context of our exercise, we calculated \( \ln(0.6) \) which equals approximately -0.511.
Because of this negative logarithmic value, it indicates that if \( e \) is raised to a power of about -0.511, it will result in a value close to 0.6. This approximation allows us to compute the value of \( x \) with a level of precision suitable for practical purposes.

It's important to remember:

• Decimal approximations are rounded to ensure precision and ease of use, particularly when needing to express results to three decimal places.

Thanks to these calculations, we can return a neatly rounded value of \( x \), ensuring that the outcome is both useful and concise, as shown in the final conversion equation:
\( x = -1.956(\ln{(\frac{y}{4.5})}) \).