Problem 26 - Introductory and Intermediate Algebra for College Students 4th

Question

Solve each exponential equation by taking the logarithm on both sides. Express the solution set in terms of logarithms. Then use a calculator to obtain a decimal approximation, correct to two decimal places, for the solution. $$9 e^{x}=99$$

Step-by-Step Solution

Answer

In terms of logarithms, the solution to the equation is \(x = ln(11)\). The decimal approximation of this solution is \(x \approx 2.40\).

Step 1: Applying Natural Logarithm

Apply the natural logarithm to both sides of the equation \(9 e^{x} = 99\). This allows to make use of the property of logarithms, where the exponent can be brought down as a multiplier. Thus it becomes \(ln(9) + ln(e^x) = ln(99)\). The logarithmic form of \(e^x\) is simply \(x\), so the equation reduces to \(ln(9) + x = ln(99)\).

Step 2: Isolate the Variable

Isolate the variable \(x\) on one side, to achieve a formula for \(x\) in terms of natural logarithms. From earlier it is known that \(ln(9) + x = ln(99)\), so \(x = ln(99) - ln(9)\). Using the logarithmic relationship \(ln(a)-ln(b) = ln(a/b)\), this expression can further be simplified to \(x = ln(99/9)\).

Step 3: Calculate Decimal Approximation

You can now replace x in the equation with the simplified logarithmic expression. Then use a calculator to find the numerical value of the natural logarithm. It is \(x = ln(11) \approx 2.40\).

Key Concepts

Natural Logarithms

When solving exponential equations, natural logarithms play a crucial role. A natural logarithm, denoted as \( \ln \), is a logarithm with the base of Euler's number \( e \), which is approximately 2.718. In the exercise, applying the natural logarithm to both sides of the equation allows us to utilize logarithmic properties conveniently. This step is important as it helps in handling the variable involved in the exponential function.
By taking the natural logarithm of both sides:

• Equation becomes easier to manipulate.
• Exponents in equations can be handled effectively.

In the equation \( 9 e^{x} = 99 \), taking \( \ln \) of both sides gives us \( \ln(9 e^{x}) = \ln(99) \). Here, logarithmic rules help in further simplification, ultimately leading us closer to solving for \( x \).

Logarithmic Properties

Logarithmic properties are mathematical tools that help to simplify logarithmic expressions. In this exercise, a key property utilized is that the logarithm of a product is the sum of logarithms: \( \ln(a \cdot b) = \ln(a) + \ln(b) \). This is used in the equation \( \ln(9 e^{x}) = \ln(99) \), which simplifies to \( \ln(9) + \ln(e^{x}) = \ln(99) \).
Another vital property is that \( \ln(e^x) = x \), effectively eliminating the \( e \) term from the equation. This property helps simplify to \( \ln(9) + x = \ln(99) \). You can also use the quotient property of logarithms:

• \( \ln(a) - \ln(b) = \ln(\frac{a}{b}) \)

This allows for the reduction \( x = \ln(99) - \ln(9) \) to \( x = \ln(\frac{99}{9}) \), or more simply \( x = \ln(11) \).
Understanding these properties is critical to efficiently manipulate and solve logarithmic equations.

Decimal Approximation

After expressing \( x \) in terms of natural logarithms, obtaining a decimal approximation is often the final step. Calculators generally cannot handle logarithm expressions directly, so approximation facilitates practical use of the result. In this exercise, once we have \( x = \ln(11) \), a calculator is used to approximate the numerical value to two decimal places.
This converts the logarithmic result into a familiar numerical standard. Using a scientific calculator:

• Enter the natural logarithm expression.
• The result, \( \ln(11) \), approximately equals 2.40.

It simplifies the reading and application of results, essential when interpreting and communicating mathematical solutions. Ultimately, decimal approximations make the derived solutions more tangible for daily applications and further numerical computations.