Problem 26 - Algebra and Trigonometry

Question

Solve each exponential equation . Express the solution set in terms of natural logarithms or common logarithms. Then use a calculator to obtain a decimal approximation, correct to two decimal places, for the solution. $$ e^{x}=0.83 $$

Step-by-Step Solution

Answer

The solution is \( x \approx -0.19 \).

Step 1: Apply the natural logarithm

Applying the natural logarithm to both sides of the equation allows unlinking the value of \( x \) from the exponent. This is done by writing the equation as: \( ln(e^{x}) = ln(0.83) \). Using the property of logarithms \( ln(a^{b}) = b * ln(a) \), the left-hand side of the equation can be simplified to \( x * ln(e) \). Since the natural logarithm of \( e \) is 1, \( x * ln(e) \) equals \( x \). This gives us the equation \( x = ln(0.83) \).

Step 2: Calculate the natural logarithm

By using a calculator, calculate the natural logarithm of 0.83 to obtain a decimal approximation for \( x \). This leaves us with: \( x \approx -0.19 \).

Step 3: Check the solution

Finally, substitute \( x \) into the original equation \( e^{x} = 0.83 \) to check the solution. If substituting \( x = -0.19 \) into the equation gives a true equality, this confirms that the solution is correct.

Key Concepts

Understanding the Natural Logarithm

The natural logarithm is a fundamental concept often abbreviated as "ln." It's used in exponential equations to find exponents, like in the equation \( e^x = 0.83 \). Here, "e" is a mathematical constant approximately equal to 2.71828. This special logarithm is significant because it simplifies calculations involving the base \( e \).
To solve an exponential equation like \( e^x = 0.83 \), we apply the natural logarithm to both sides. This technique helps "unpack" the exponent. So, we write:

• \( \ln(e^x) = \ln(0.83) \)

Using the property of logarithms, \( \ln(a^b) = b \ln(a) \), the left-side simplifies to \( x \ln(e) \). Since \( \ln(e) = 1 \), this becomes simply \( x = \ln(0.83) \).
In other words, the natural log translates the equation from an exponent form to a more straightforward linear form, making it much easier to solve for \( x \).

Obtaining a Decimal Approximation

After expressing the solution using natural logarithms, the next task is to find a decimal approximation for the numerical value, which offers a more tangible solution. In this case, we need to calculate \( \ln(0.83) \) to two decimal places.
A scientific calculator or a logarithmic function on most calculators can help us find \( \ln(0.83) \). When calculated, it comes out as approximately \( -0.186 \). When rounded to two decimal places, it becomes \( -0.19 \).
These approximations are helpful because they let us work with numbers that are easier to understand and use, particularly in practical scenarios. The precise decimal generated provides clarity when verifying or interpreting the results.

Exploring Logarithmic Properties

Logarithmic properties are key to manipulating and understanding equations that involve logarithms. Knowing these properties allows you to solve equations more easily.

• Power Rule: \( \ln(a^b) = b \cdot \ln(a) \) helps convert exponential expressions into products.
• Logarithm of 1: Any log of 1, for any base, is always 0, i.e., \( \ln(1) = 0 \).
• Product Rule: \( \ln(ab) = \ln(a) + \ln(b) \) breaks down the log of a product into the sum of logs.

These rules simplify complex expressions and are particularly useful in solving equations where the variable is in the exponent, as seen in our problem \( e^x = 0.83 \).
Recognizing how to simplify the equation using properties like the Power Rule can transform an exponential equation into something much more manageable. Ultimately, understanding these properties deepens comprehension and boosts capability in handling a wide array of logarithmic and exponential problems.